Mixing
What is the smallest $k$ , such that $44^{44}+k$ has the desired property?
I search the smallest positive integer $k$, such that $44^{44}+k$ splits into three distinct prime factors each having $25$ decimal digits.
The $21$-digit number $k=621725397145122340237$ does the job, but it is hard to imagine that there are no smaller solutions.
Enzo Creti found the very near miss :
$$44^{44}+202693 = P24cdot P25cdot P25$$
Is there any better way than just factoring all numbers until the desired number is found ?
number-theory elementary-number-theory prime-factorization
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
add a comment |
I search the smallest positive integer $k$, such that $44^{44}+k$ splits into three distinct prime factors each having $25$ decimal digits.
The $21$-digit number $k=621725397145122340237$ does the job, but it is hard to imagine that there are no smaller solutions.
Enzo Creti found the very near miss :
$$44^{44}+202693 = P24cdot P25cdot P25$$
Is there any better way than just factoring all numbers until the desired number is found ?
number-theory elementary-number-theory prime-factorization
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
1
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09
add a comment |
I search the smallest positive integer $k$, such that $44^{44}+k$ splits into three distinct prime factors each having $25$ decimal digits.
The $21$-digit number $k=621725397145122340237$ does the job, but it is hard to imagine that there are no smaller solutions.
Enzo Creti found the very near miss :
$$44^{44}+202693 = P24cdot P25cdot P25$$
Is there any better way than just factoring all numbers until the desired number is found ?
number-theory elementary-number-theory prime-factorization
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
I search the smallest positive integer $k$, such that $44^{44}+k$ splits into three distinct prime factors each having $25$ decimal digits.
The $21$-digit number $k=621725397145122340237$ does the job, but it is hard to imagine that there are no smaller solutions.
Enzo Creti found the very near miss :
$$44^{44}+202693 = P24cdot P25cdot P25$$
Is there any better way than just factoring all numbers until the desired number is found ?
number-theory elementary-number-theory prime-factorization
number-theory elementary-number-theory prime-factorization
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
edited Nov 21 '18 at 11:25
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
asked Nov 21 '18 at 10:08
Peter
46.7k1039125
46.7k1039125
1
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09
add a comment |
1
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09
1
1
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09
add a comment |
1 Answer
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To find n primes closest to $a^a$ it is reasonable to consider closest primes to nth root of $a^a$.Example; Find three distinct primes such that their product is closest to number $11^{11}$:
$11^{11/3}≈6636$
We check this:Three consecutive primes close to this number are $6581$, $6599$ and $6607$ and we have:
$6581^3-11^{11}=-291449670$
$6599^3-11^{11}=2053669188$
$6607^3-11^{11}=3100059932$
$6581times 6599times 6607-11^{11}=1617250922$
$6581times 6599times 6607=11^{11}+1617250922$
Now we try to find more optimized value. This question is somehow related to following optimization problem:
Find the dimensions of a rectangle cube inscribed in sphere with radius R and having maximum volume. I assumed the base of cube is square and found $AB=1.4 h$ where AB is the side of square and h is the height of cube(Calculations can be posted in a separate question if there are interests in it).Now we have a prime $p_1=h$ and two other primes $p_2=1.4 p_1$ and $p_3$ close to $p_2$. The mean value of theses three primes is:
$frac{1.4 p_1+1.4p_1+p_1}{3}≈1.26 p_1$
Let's take this mean value equal to $11^{11/3}≈6636$ then $p_1=6636/1.26≈5266$. The closest prime to this number is $5261$, let $p_1=5261$, then:
$p_2=5261times 1.4=7365$
The closest prime to this number is $7369$, let $p_2=7369$. The next prime to this is $7393$, let $p_3=7393$, then we have:
$5261*7369*7393-11^{11}=1302437826$
As can be seen the value of $k=1302437826$ is smaller than previous value we found.There may be even better optimization which defines the ratio between the primes.
Since we have to find three distinct primes, they must be close to the third root of $44^{44}$.My computer gives:
$44^{frac{44}{3}}≈26.6times 10^{23}$
So we can take this value as the mean of primes and follow the above method.
answered Nov 22 '18 at 20:13
sirous
1,6091513
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
add a comment |
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1 Answer
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1 Answer
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To find n primes closest to $a^a$ it is reasonable to consider closest primes to nth root of $a^a$.Example; Find three distinct primes such that their product is closest to number $11^{11}$:
$11^{11/3}≈6636$
We check this:Three consecutive primes close to this number are $6581$, $6599$ and $6607$ and we have:
$6581^3-11^{11}=-291449670$
$6599^3-11^{11}=2053669188$
$6607^3-11^{11}=3100059932$
$6581times 6599times 6607-11^{11}=1617250922$
$6581times 6599times 6607=11^{11}+1617250922$
Now we try to find more optimized value. This question is somehow related to following optimization problem:
Find the dimensions of a rectangle cube inscribed in sphere with radius R and having maximum volume. I assumed the base of cube is square and found $AB=1.4 h$ where AB is the side of square and h is the height of cube(Calculations can be posted in a separate question if there are interests in it).Now we have a prime $p_1=h$ and two other primes $p_2=1.4 p_1$ and $p_3$ close to $p_2$. The mean value of theses three primes is:
$frac{1.4 p_1+1.4p_1+p_1}{3}≈1.26 p_1$
Let's take this mean value equal to $11^{11/3}≈6636$ then $p_1=6636/1.26≈5266$. The closest prime to this number is $5261$, let $p_1=5261$, then:
$p_2=5261times 1.4=7365$
The closest prime to this number is $7369$, let $p_2=7369$. The next prime to this is $7393$, let $p_3=7393$, then we have:
$5261*7369*7393-11^{11}=1302437826$
As can be seen the value of $k=1302437826$ is smaller than previous value we found.There may be even better optimization which defines the ratio between the primes.
Since we have to find three distinct primes, they must be close to the third root of $44^{44}$.My computer gives:
$44^{frac{44}{3}}≈26.6times 10^{23}$
So we can take this value as the mean of primes and follow the above method.
answered Nov 22 '18 at 20:13
sirous
1,6091513
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
add a comment |
To find n primes closest to $a^a$ it is reasonable to consider closest primes to nth root of $a^a$.Example; Find three distinct primes such that their product is closest to number $11^{11}$:
$11^{11/3}≈6636$
We check this:Three consecutive primes close to this number are $6581$, $6599$ and $6607$ and we have:
$6581^3-11^{11}=-291449670$
$6599^3-11^{11}=2053669188$
$6607^3-11^{11}=3100059932$
$6581times 6599times 6607-11^{11}=1617250922$
$6581times 6599times 6607=11^{11}+1617250922$
Now we try to find more optimized value. This question is somehow related to following optimization problem:
Find the dimensions of a rectangle cube inscribed in sphere with radius R and having maximum volume. I assumed the base of cube is square and found $AB=1.4 h$ where AB is the side of square and h is the height of cube(Calculations can be posted in a separate question if there are interests in it).Now we have a prime $p_1=h$ and two other primes $p_2=1.4 p_1$ and $p_3$ close to $p_2$. The mean value of theses three primes is:
$frac{1.4 p_1+1.4p_1+p_1}{3}≈1.26 p_1$
Let's take this mean value equal to $11^{11/3}≈6636$ then $p_1=6636/1.26≈5266$. The closest prime to this number is $5261$, let $p_1=5261$, then:
$p_2=5261times 1.4=7365$
The closest prime to this number is $7369$, let $p_2=7369$. The next prime to this is $7393$, let $p_3=7393$, then we have:
$5261*7369*7393-11^{11}=1302437826$
As can be seen the value of $k=1302437826$ is smaller than previous value we found.There may be even better optimization which defines the ratio between the primes.
Since we have to find three distinct primes, they must be close to the third root of $44^{44}$.My computer gives:
$44^{frac{44}{3}}≈26.6times 10^{23}$
So we can take this value as the mean of primes and follow the above method.
answered Nov 22 '18 at 20:13
sirous
1,6091513
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
add a comment |
To find n primes closest to $a^a$ it is reasonable to consider closest primes to nth root of $a^a$.Example; Find three distinct primes such that their product is closest to number $11^{11}$:
$11^{11/3}≈6636$
We check this:Three consecutive primes close to this number are $6581$, $6599$ and $6607$ and we have:
$6581^3-11^{11}=-291449670$
$6599^3-11^{11}=2053669188$
$6607^3-11^{11}=3100059932$
$6581times 6599times 6607-11^{11}=1617250922$
$6581times 6599times 6607=11^{11}+1617250922$
Now we try to find more optimized value. This question is somehow related to following optimization problem:
Find the dimensions of a rectangle cube inscribed in sphere with radius R and having maximum volume. I assumed the base of cube is square and found $AB=1.4 h$ where AB is the side of square and h is the height of cube(Calculations can be posted in a separate question if there are interests in it).Now we have a prime $p_1=h$ and two other primes $p_2=1.4 p_1$ and $p_3$ close to $p_2$. The mean value of theses three primes is:
$frac{1.4 p_1+1.4p_1+p_1}{3}≈1.26 p_1$
Let's take this mean value equal to $11^{11/3}≈6636$ then $p_1=6636/1.26≈5266$. The closest prime to this number is $5261$, let $p_1=5261$, then:
$p_2=5261times 1.4=7365$
The closest prime to this number is $7369$, let $p_2=7369$. The next prime to this is $7393$, let $p_3=7393$, then we have:
$5261*7369*7393-11^{11}=1302437826$
As can be seen the value of $k=1302437826$ is smaller than previous value we found.There may be even better optimization which defines the ratio between the primes.
Since we have to find three distinct primes, they must be close to the third root of $44^{44}$.My computer gives:
$44^{frac{44}{3}}≈26.6times 10^{23}$
So we can take this value as the mean of primes and follow the above method.
answered Nov 22 '18 at 20:13
sirous
1,6091513
To find n primes closest to $a^a$ it is reasonable to consider closest primes to nth root of $a^a$.Example; Find three distinct primes such that their product is closest to number $11^{11}$:
$11^{11/3}≈6636$
We check this:Three consecutive primes close to this number are $6581$, $6599$ and $6607$ and we have:
$6581^3-11^{11}=-291449670$
$6599^3-11^{11}=2053669188$
$6607^3-11^{11}=3100059932$
$6581times 6599times 6607-11^{11}=1617250922$
$6581times 6599times 6607=11^{11}+1617250922$
Now we try to find more optimized value. This question is somehow related to following optimization problem:
Find the dimensions of a rectangle cube inscribed in sphere with radius R and having maximum volume. I assumed the base of cube is square and found $AB=1.4 h$ where AB is the side of square and h is the height of cube(Calculations can be posted in a separate question if there are interests in it).Now we have a prime $p_1=h$ and two other primes $p_2=1.4 p_1$ and $p_3$ close to $p_2$. The mean value of theses three primes is:
$frac{1.4 p_1+1.4p_1+p_1}{3}≈1.26 p_1$
Let's take this mean value equal to $11^{11/3}≈6636$ then $p_1=6636/1.26≈5266$. The closest prime to this number is $5261$, let $p_1=5261$, then:
$p_2=5261times 1.4=7365$
The closest prime to this number is $7369$, let $p_2=7369$. The next prime to this is $7393$, let $p_3=7393$, then we have:
$5261*7369*7393-11^{11}=1302437826$
As can be seen the value of $k=1302437826$ is smaller than previous value we found.There may be even better optimization which defines the ratio between the primes.
Since we have to find three distinct primes, they must be close to the third root of $44^{44}$.My computer gives:
$44^{frac{44}{3}}≈26.6times 10^{23}$
So we can take this value as the mean of primes and follow the above method.
answered Nov 22 '18 at 20:13
sirous
1,6091513
edited Nov 23 '18 at 16:43
answered Nov 22 '18 at 20:13
sirous
1,6091513
answered Nov 22 '18 at 20:13
sirous
1,6091513
answered Nov 22 '18 at 20:13
sirous
1,6091513
1,6091513
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
add a comment |
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
1
1
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
Instead of three primes very close to $44^{44/3}$, I could imagine that (for example) two primes much smaller* than $44^{44/3}$ and one much larger* than $44^{44/3}$ could make a product much closer to $44^{44}$. (*All primes still having 25 digits of course.)
– alex.jordan
Nov 22 '18 at 23:09
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
@alex.jordan, I tried all these. Can you give an example for four digits primes closest to $11^11$. If it is better than my solution then I accept your claim.
– sirous
Nov 23 '18 at 5:28
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
For example, with $21436$ replacing $44^{44}$, the cube root is about $28$. Nearby primes include $19,23,29,31$. No product of these or other nearby primes does a better job than $13cdot17cdot97$, which is $21437$, only one more. So here we see two primes much smaller than the cube root and one much larger, doing a better job than three close to the cube root.
– alex.jordan
Nov 23 '18 at 19:36
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1
Maybe start in the other end? Multiply 25-digit primes so that you get something marginally larger than $44^{44}$. I don't know whether that is in any way easier.
– Arthur
Nov 21 '18 at 10:13
@Arthur I constructed such numbers, but they are too faw away from $44^{44}$
– Peter
Nov 21 '18 at 10:14
Hi guys, what does $P24$ and $P25$ mean?
– Lee
Nov 21 '18 at 11:20
@Lee $P24$ just means a prime factor with $24$ decimal digits, analogue $P25$
– Peter
Nov 21 '18 at 11:28
I can imagine coming up with a method that could nicely give some $k$ that is small but not necessarily minimal. But it seems like to find the minimal $k$, you will have to do some sort of brute force factorizing or sieving of the integers that immediately follow $44^{44}$. If you end up doing that anyway, I wonder if it's just as good to start with that. So with a computer, start factorizing/sieving $44^{44}+1, 44^{44}+2,ldots$. When I say "sieving", that includes skipping over even numbers and other things like that.
– alex.jordan
Nov 24 '18 at 1:09