Mixing
Determining the missing digits of $15! equiv 1square0767436square000$ without actually calculating the...
$begingroup$
$$15! equiv 1cdot 2cdot 3cdot,cdots,cdot 15 equiv 1square0767436square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! approx sqrt{2picdot 15}
cdot left(frac{15}{e}right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 mod phantom{1}9 equiv ,34 + d_0 + d_1 mod phantom{1}9 equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 mod 11 equiv d_0 + d_1 mod 11 equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
elementary-number-theory factorial
asked Jan 2 at 21:30
DarkiceDarkice
1415
$endgroup$
|
show 3 more comments
$begingroup$
$$15! equiv 1cdot 2cdot 3cdot,cdots,cdot 15 equiv 1square0767436square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! approx sqrt{2picdot 15}
cdot left(frac{15}{e}right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 mod phantom{1}9 equiv ,34 + d_0 + d_1 mod phantom{1}9 equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 mod 11 equiv d_0 + d_1 mod 11 equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
elementary-number-theory factorial
asked Jan 2 at 21:30
DarkiceDarkice
1415
$endgroup$
8
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
3
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
1
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
6
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09
|
show 3 more comments
$begingroup$
$$15! equiv 1cdot 2cdot 3cdot,cdots,cdot 15 equiv 1square0767436square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! approx sqrt{2picdot 15}
cdot left(frac{15}{e}right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 mod phantom{1}9 equiv ,34 + d_0 + d_1 mod phantom{1}9 equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 mod 11 equiv d_0 + d_1 mod 11 equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
elementary-number-theory factorial
asked Jan 2 at 21:30
DarkiceDarkice
1415
$endgroup$
$$15! equiv 1cdot 2cdot 3cdot,cdots,cdot 15 equiv 1square0767436square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! approx sqrt{2picdot 15}
cdot left(frac{15}{e}right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 mod phantom{1}9 equiv ,34 + d_0 + d_1 mod phantom{1}9 equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 mod 11 equiv d_0 + d_1 mod 11 equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
elementary-number-theory factorial
elementary-number-theory factorial
asked Jan 2 at 21:30
DarkiceDarkice
1415
asked Jan 2 at 21:30
DarkiceDarkice
1415
edited Jan 2 at 23:48
user630758
asked Jan 2 at 21:30
DarkiceDarkice
1415
asked Jan 2 at 21:30
DarkiceDarkice
1415
asked Jan 2 at 21:30
DarkiceDarkice
1415
1415
8
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
3
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
1
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
6
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09
|
show 3 more comments
8
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
3
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
1
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
6
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09
8
8
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
3
3
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
1
1
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
6
6
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09
|
show 3 more comments
6 Answers
6
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votes
$begingroup$
Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, the final $square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $square$ is an $8$. Casting out $9$'s now reveals that the first $square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
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add a comment |
$begingroup$
You can cast out $9$’s and $11$’s:
begin{align}
1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \
1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y
end{align}
Thus $x+y=11$ (it can't be $x=y=0$).
Then find the remainder modulo $10000$; since
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.
answered Jan 2 at 23:49
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 mod 7 equiv 0$
answered Jan 2 at 23:15
DarkiceDarkice
1415
$endgroup$
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
add a comment |
$begingroup$
Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b equiv b pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$
$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0le a le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
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$begingroup$
Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
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$begingroup$
$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
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6 Answers
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6 Answers
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$begingroup$
Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, the final $square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $square$ is an $8$. Casting out $9$'s now reveals that the first $square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
$endgroup$
add a comment |
$begingroup$
Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, the final $square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $square$ is an $8$. Casting out $9$'s now reveals that the first $square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
$endgroup$
add a comment |
$begingroup$
Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, the final $square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $square$ is an $8$. Casting out $9$'s now reveals that the first $square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
$endgroup$
Another way to reason is to note that $15!$ is divisible by $2cdot4cdot2cdot8cdot2cdot4cdot2=2^{11}$, which means $1square0767436square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8mid1000$ and $8mid360$, the final $square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $square$ is an $8$. Casting out $9$'s now reveals that the first $square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
answered Jan 2 at 23:47
Barry CipraBarry Cipra
59.4k653125
59.4k653125
add a comment |
add a comment |
$begingroup$
You can cast out $9$’s and $11$’s:
begin{align}
1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \
1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y
end{align}
Thus $x+y=11$ (it can't be $x=y=0$).
Then find the remainder modulo $10000$; since
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.
answered Jan 2 at 23:49
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
You can cast out $9$’s and $11$’s:
begin{align}
1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \
1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y
end{align}
Thus $x+y=11$ (it can't be $x=y=0$).
Then find the remainder modulo $10000$; since
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.
answered Jan 2 at 23:49
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
You can cast out $9$’s and $11$’s:
begin{align}
1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \
1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y
end{align}
Thus $x+y=11$ (it can't be $x=y=0$).
Then find the remainder modulo $10000$; since
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.
answered Jan 2 at 23:49
egregegreg
180k1485202
$endgroup$
You can cast out $9$’s and $11$’s:
begin{align}
1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \
1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y
end{align}
Thus $x+y=11$ (it can't be $x=y=0$).
Then find the remainder modulo $10000$; since
$$
15!=2^{11}cdot 3^6cdot 5^3cdot 7^2cdot11cdot13=1000cdot 2^8cdot3^6cdot7^2cdot 11cdot 13
$$
this means finding the remainder modulo $10$ of
$$
2^8cdot3^6cdot7^2cdot 11cdot 13
$$
that gives $8$ with a short computation.
answered Jan 2 at 23:49
egregegreg
180k1485202
answered Jan 2 at 23:49
egregegreg
180k1485202
answered Jan 2 at 23:49
egregegreg
180k1485202
answered Jan 2 at 23:49
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 mod 7 equiv 0$
answered Jan 2 at 23:15
DarkiceDarkice
1415
$endgroup$
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
add a comment |
$begingroup$
Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 mod 7 equiv 0$
answered Jan 2 at 23:15
DarkiceDarkice
1415
$endgroup$
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
add a comment |
$begingroup$
Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 mod 7 equiv 0$
answered Jan 2 at 23:15
DarkiceDarkice
1415
$endgroup$
Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 mod 7 equiv 0$
answered Jan 2 at 23:15
DarkiceDarkice
1415
answered Jan 2 at 23:15
DarkiceDarkice
1415
answered Jan 2 at 23:15
DarkiceDarkice
1415
answered Jan 2 at 23:15
DarkiceDarkice
1415
1415
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
add a comment |
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side.
$endgroup$
– no0ob
Jan 2 at 23:34
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
$begingroup$
The order is also given by the divisibilty rule for 7.
$endgroup$
– Darkice
Jan 3 at 8:54
add a comment |
$begingroup$
Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b equiv b pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$
$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0le a le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
$endgroup$
add a comment |
$begingroup$
Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b equiv b pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$
$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0le a le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
$endgroup$
add a comment |
$begingroup$
Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b equiv b pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$
$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0le a le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
$endgroup$
Okay, $15! = 1*2*3..... *15=1a0767436b000$.
Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.
If we divide $15!$ by $100 = 8*5^3$ we get
$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$
If we want to find the last digit of that we can do
$1a0767436b equiv b pmod {10}$ and
$1*2*3*4*6*7*9*2*11*12*13*14*3equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3equiv$
$-2^9*3^4 equiv -512*81equiv -2 equiv 8pmod {10}$..
So $b = 8$.
But what is $a$?
Well, $11|1a0767436b$ and $9|1a0767436b$.
So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.
So $-a -8 =11k$ so as $0le a le 9$ we have $a = 3$.
And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.
We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
answered Jan 3 at 0:53
fleabloodfleablood
68.8k22685
68.8k22685
add a comment |
add a comment |
$begingroup$
Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
$endgroup$
Let $d_1$ and $d_2$ be the two unknown digits.
The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.
$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.
Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
answered Jan 3 at 4:55
farruhotafarruhota
19.7k2738
19.7k2738
add a comment |
add a comment |
$begingroup$
$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$15!=2^{11}cdot 5^3cdot 7^2cdot 11cdot 13=(1000)X$ where $X=2^8cdot 3^6cdot 7^2cdot 11cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6cdot 9cdot 9 cdot 1cdot 3equiv 6cdot(-1)^2cdot 3equiv 18equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 10:16
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
add a comment |
add a comment |
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8
$begingroup$
Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even.
$endgroup$
– lulu
Jan 2 at 21:36
$begingroup$
Why is the sum of the digits divisible by 9?
$endgroup$
– Darkice
Jan 2 at 21:37
3
$begingroup$
@Darkice because for every natural number $ngeq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine.
$endgroup$
– JMoravitz
Jan 2 at 21:41
1
$begingroup$
Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements.
$endgroup$
– Darkice
Jan 2 at 21:59
6
$begingroup$
For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}cdot3^6cdot5^3cdot7^2cdot11cdot13$ the final nonzero digit will be $2^8cdot3^6cdot7^2cdot11cdot13pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8cdot3^6cdot7^2cdot11cdot13equiv 6cdot 9cdot 9cdot 1cdot 3equiv 8pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes.
$endgroup$
– JMoravitz
Jan 2 at 22:09