Mixing
Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number?
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This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
$endgroup$
add a comment |
$begingroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
$endgroup$
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
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– Jeppe Stig Nielsen
Jan 29 at 21:17
3
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This type of number is called a “repunit”.
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– Dan
Jan 29 at 21:21
1
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@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05
add a comment |
$begingroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
$endgroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
edited Jan 30 at 7:37
Asaf Karagila♦
307k33439772
307k33439772
asked Jan 29 at 6:34
user69284user69284
17918
asked Jan 29 at 6:34
user69284user69284
17918
asked Jan 29 at 6:34
user69284user69284
17918
17918
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
Jan 29 at 21:17
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
Jan 29 at 21:21
1
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05
add a comment |
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
Jan 29 at 21:17
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
Jan 29 at 21:21
1
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43
1
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
Jan 29 at 21:17
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
Jan 29 at 21:17
3
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
Jan 29 at 21:21
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
Jan 29 at 21:21
1
1
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
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1
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Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
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– user69284
Jan 29 at 6:45
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Check the edit. I have found a better (and easier to understand) solution.
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– Ekesh Kumar
Jan 29 at 6:51
12
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"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
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– Chris
Jan 29 at 10:20
19
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Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
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– Chris
Jan 29 at 10:22
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
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– boboquack
Jan 29 at 22:43
|
show 1 more comment
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You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered Jan 29 at 10:25
ChrisChris
4481513
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1
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If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
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– potestasity
Jan 29 at 12:57
6
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@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
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– David Robinson
Jan 29 at 14:04
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@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
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– Chris
Jan 29 at 14:17
2
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@Chris If you had 55 2's you simply would divide by 2.
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– rus9384
Jan 29 at 14:36
1
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@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
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– Random832
Jan 29 at 14:36
|
show 6 more comments
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More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered Jan 29 at 6:49
heropupheropup
64.8k764103
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1
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Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
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– orion
Jan 30 at 7:55
add a comment |
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This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered Jan 29 at 14:50
DanijelDanijel
836518
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If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
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– gnasher729
Jan 29 at 18:46
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I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
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– Acccumulation
Jan 29 at 21:05
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This is the best answer here imo
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– goblin
Jan 30 at 3:26
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@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
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– Danijel
Jan 30 at 9:13
add a comment |
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The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
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6
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Can you clarify how you obtained this?
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– Pedro A
Jan 29 at 11:14
3
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I think if the OP searched for the factorization alone, he had used a calculator in the first place
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– Nico Haase
Jan 29 at 14:14
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@PedroA Several repunit factorizations are available here
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– LegionMammal978
Jan 29 at 15:33
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I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
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– Daniel Schepler
Jan 29 at 18:10
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Or Maple or any other computer algebra system.
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– David G. Stork
Jan 29 at 18:10
add a comment |
$begingroup$
You can either do five blocks of eleven 1s:
$$11111111111 11111111111 11111111111 11111111111 11111111111 \ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
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add a comment |
Your Answer
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
|
show 1 more comment
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
|
show 1 more comment
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
edited Jan 29 at 21:24
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
answered Jan 29 at 6:39
Ekesh KumarEkesh Kumar
1,08428
1,08428
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
|
show 1 more comment
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
1
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
Jan 29 at 6:45
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
Jan 29 at 6:51
12
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
Jan 29 at 10:20
19
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
Jan 29 at 10:22
1
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
Jan 29 at 22:43
|
show 1 more comment
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered Jan 29 at 10:25
ChrisChris
4481513
$endgroup$
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
6
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
|
show 6 more comments
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered Jan 29 at 10:25
ChrisChris
4481513
$endgroup$
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
6
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
|
show 6 more comments
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered Jan 29 at 10:25
ChrisChris
4481513
$endgroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered Jan 29 at 10:25
ChrisChris
4481513
answered Jan 29 at 10:25
ChrisChris
4481513
answered Jan 29 at 10:25
ChrisChris
4481513
answered Jan 29 at 10:25
ChrisChris
4481513
4481513
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
6
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
|
show 6 more comments
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
6
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
1
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
Jan 29 at 12:57
6
6
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
Jan 29 at 14:04
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
Jan 29 at 14:17
2
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
Jan 29 at 14:36
1
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
Jan 29 at 14:36
|
show 6 more comments
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered Jan 29 at 6:49
heropupheropup
64.8k764103
$endgroup$
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
add a comment |
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered Jan 29 at 6:49
heropupheropup
64.8k764103
$endgroup$
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
add a comment |
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered Jan 29 at 6:49
heropupheropup
64.8k764103
$endgroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered Jan 29 at 6:49
heropupheropup
64.8k764103
answered Jan 29 at 6:49
heropupheropup
64.8k764103
answered Jan 29 at 6:49
heropupheropup
64.8k764103
answered Jan 29 at 6:49
heropupheropup
64.8k764103
64.8k764103
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
add a comment |
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
1
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
Jan 30 at 7:55
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered Jan 29 at 14:50
DanijelDanijel
836518
$endgroup$
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered Jan 29 at 14:50
DanijelDanijel
836518
$endgroup$
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered Jan 29 at 14:50
DanijelDanijel
836518
$endgroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered Jan 29 at 14:50
DanijelDanijel
836518
answered Jan 29 at 14:50
DanijelDanijel
836518
answered Jan 29 at 14:50
DanijelDanijel
836518
answered Jan 29 at 14:50
DanijelDanijel
836518
836518
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
add a comment |
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
Jan 29 at 18:46
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
Jan 29 at 21:05
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
Jan 30 at 3:26
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
Jan 30 at 9:13
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
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6
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Can you clarify how you obtained this?
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– Pedro A
Jan 29 at 11:14
3
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I think if the OP searched for the factorization alone, he had used a calculator in the first place
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– Nico Haase
Jan 29 at 14:14
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@PedroA Several repunit factorizations are available here
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– LegionMammal978
Jan 29 at 15:33
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I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
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– Daniel Schepler
Jan 29 at 18:10
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Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
Jan 29 at 11:14
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
Jan 29 at 14:14
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
Jan 29 at 15:33
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
Jan 29 at 18:10
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
$endgroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
edited Jan 29 at 18:11
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
answered Jan 29 at 6:59
David G. StorkDavid G. Stork
11.6k41533
11.6k41533
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
Jan 29 at 11:14
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
Jan 29 at 14:14
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
Jan 29 at 15:33
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
Jan 29 at 18:10
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
add a comment |
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
Jan 29 at 11:14
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
Jan 29 at 14:14
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
Jan 29 at 15:33
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
Jan 29 at 18:10
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
6
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
Jan 29 at 11:14
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
Jan 29 at 11:14
3
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
Jan 29 at 14:14
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
Jan 29 at 14:14
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
Jan 29 at 15:33
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
Jan 29 at 15:33
$begingroup$
I got the same thing just entering
factor((10^55-1)/9); into Maxima.$endgroup$
– Daniel Schepler
Jan 29 at 18:10
$begingroup$
I got the same thing just entering
factor((10^55-1)/9); into Maxima.$endgroup$
– Daniel Schepler
Jan 29 at 18:10
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
Jan 29 at 18:10
add a comment |
$begingroup$
You can either do five blocks of eleven 1s:
$$11111111111 11111111111 11111111111 11111111111 11111111111 \ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
$endgroup$
add a comment |
$begingroup$
You can either do five blocks of eleven 1s:
$$11111111111 11111111111 11111111111 11111111111 11111111111 \ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
$endgroup$
add a comment |
$begingroup$
You can either do five blocks of eleven 1s:
$$11111111111 11111111111 11111111111 11111111111 11111111111 \ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
$endgroup$
You can either do five blocks of eleven 1s:
$$11111111111 11111111111 11111111111 11111111111 11111111111 \ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
answered Feb 6 at 18:38
community wiki
Tanner Swett
add a comment |
add a comment |
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$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
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– Ekesh Kumar
Jan 29 at 6:43
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
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– Jeppe Stig Nielsen
Jan 29 at 21:17
3
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This type of number is called a “repunit”.
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– Dan
Jan 29 at 21:21
1
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@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
Jan 30 at 9:05