Positive divisors of n = $2^{14} cdot 3^9 cdot 5^8 cdot 7^{10} cdot 11^3 cdot 13^5 cdot 37^{10}$












1












$begingroup$


How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2



The answer is (1)(1)(2)(2)(1)(1)(3) = 12



I don't understand though because I would have done something like:



2: [(14-10)/3]+1 = 2 (taking the floor)



3: [(9-9)/3]+1 = 1



5: [(8-2)/3]+1 = 3



7: [(10-5)/3]+1 = 2



11: [(3-2)/3]+1 = 1



13: [(5-2)/3]+1 = 2



37: [(10-2)/3]+1 =3



2*1*3*2*1*2*3






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:23












  • $begingroup$
    @JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:35






  • 1




    $begingroup$
    @rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:39










  • $begingroup$
    @JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:47












  • $begingroup$
    Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 22:03


















1












$begingroup$


How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2



The answer is (1)(1)(2)(2)(1)(1)(3) = 12



I don't understand though because I would have done something like:



2: [(14-10)/3]+1 = 2 (taking the floor)



3: [(9-9)/3]+1 = 1



5: [(8-2)/3]+1 = 3



7: [(10-5)/3]+1 = 2



11: [(3-2)/3]+1 = 1



13: [(5-2)/3]+1 = 2



37: [(10-2)/3]+1 =3



2*1*3*2*1*2*3






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:23












  • $begingroup$
    @JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:35






  • 1




    $begingroup$
    @rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:39










  • $begingroup$
    @JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:47












  • $begingroup$
    Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 22:03
















1












1








1





$begingroup$


How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2



The answer is (1)(1)(2)(2)(1)(1)(3) = 12



I don't understand though because I would have done something like:



2: [(14-10)/3]+1 = 2 (taking the floor)



3: [(9-9)/3]+1 = 1



5: [(8-2)/3]+1 = 3



7: [(10-5)/3]+1 = 2



11: [(3-2)/3]+1 = 1



13: [(5-2)/3]+1 = 2



37: [(10-2)/3]+1 =3



2*1*3*2*1*2*3






combinatorics







share|cite|improve this question











$endgroup$




How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2



The answer is (1)(1)(2)(2)(1)(1)(3) = 12



I don't understand though because I would have done something like:



2: [(14-10)/3]+1 = 2 (taking the floor)



3: [(9-9)/3]+1 = 1



5: [(8-2)/3]+1 = 3



7: [(10-5)/3]+1 = 2



11: [(3-2)/3]+1 = 1



13: [(5-2)/3]+1 = 2



37: [(10-2)/3]+1 =3



2*1*3*2*1*2*3






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 21:31









Aniruddh Venkatesan

161113




161113










asked Dec 8 '18 at 21:14









DevAllanPerDevAllanPer

1336




1336








  • 1




    $begingroup$
    There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:23












  • $begingroup$
    @JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:35






  • 1




    $begingroup$
    @rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:39










  • $begingroup$
    @JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:47












  • $begingroup$
    Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 22:03
















  • 1




    $begingroup$
    There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:23












  • $begingroup$
    @JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:35






  • 1




    $begingroup$
    @rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:39










  • $begingroup$
    @JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 21:47












  • $begingroup$
    Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
    $endgroup$
    – rsadhvika
    Dec 8 '18 at 22:03










1




1




$begingroup$
There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
$endgroup$
– JMoravitz
Dec 8 '18 at 21:23






$begingroup$
There is exactly one multiple of three between $10$ and $14$ (inclusive), namely $12$, which is why there is exactly one possible exponent for $2$ in the types of numbers we desire to count. Your mistaken approach seems to think there are two for some reason. While it is possible for some ranges of five consecutive integers to contain two multiples of $3$ (for example 3,4,5,6,7) it is also possible for five consecutive integers to only contain one multiple of three (for example 10,11,12,13,14), so there is more to it than simply the length of the interval in question.
$endgroup$
– JMoravitz
Dec 8 '18 at 21:23














$begingroup$
@JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
$endgroup$
– rsadhvika
Dec 8 '18 at 21:35




$begingroup$
@JMoravitz Interesting, wonder if given a random interval like $[10, 14]$, is it possible to find a formula for the exact number of multiples of $3$.
$endgroup$
– rsadhvika
Dec 8 '18 at 21:35




1




1




$begingroup$
@rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
$endgroup$
– JMoravitz
Dec 8 '18 at 21:39




$begingroup$
@rsadhvika of course, and it will depend on the lower bound's remainder when divided by 3
$endgroup$
– JMoravitz
Dec 8 '18 at 21:39












$begingroup$
@JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
$endgroup$
– rsadhvika
Dec 8 '18 at 21:47






$begingroup$
@JMoravitz for the interval $[a, b]$, something like $$leftlceil dfrac{b-a+1}{3}rightrceil + a pmod{3}$$ ? I'm not sure if it works, still testing..
$endgroup$
– rsadhvika
Dec 8 '18 at 21:47














$begingroup$
Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
$endgroup$
– rsadhvika
Dec 8 '18 at 22:03






$begingroup$
Looks I have it. Number of divisors of $n$ over the interval $[a,b]$ is $$leftlfloor dfrac{b-a+ a pmod{n}}{n}rightrfloor $$ does that look okay to you @JMoravitz
$endgroup$
– rsadhvika
Dec 8 '18 at 22:03












2 Answers
2






active

oldest

votes


















1












$begingroup$

For $d$ to be a divisor of $n$, $d$ must be of the form $2^a cdot 3^b cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $0 leq a leq 14, 0 leq b leq 9, 0 leq c leq 8, 0 leq d leq 10, 0 leq eleq 3, 0leq f leq 5, 0 leq g leq 10$.



Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a cdot 3^color{red}{9} cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $color{red}{10} leq a leq 14, color{red}{2} leq c leq 8, color{red}{5} leq d leq 10, color{red}{2} leq eleq 3, color{red}{2}leq f leq 5, color{red}{2} leq g leq 10$.



Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c in {3,6}, d in {6,9}, e=3, f=3, g in {3,6,9}.$$
Thus the total number of choices we have are
$$2 cdot 2 cdot 3=12.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
    $endgroup$
    – DevAllanPer
    Dec 8 '18 at 21:34












  • $begingroup$
    @DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
    $endgroup$
    – Anurag A
    Dec 9 '18 at 0:45





















1












$begingroup$

Since we are required to be a multiple of $2^{10}cdot3^9cdot5^2cdot7^5cdot11^2cdot 13^2cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}cdot3^9cdot5^3cdot7^6cdot11^3cdot 13^3cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:20












  • $begingroup$
    @JMoravitz Whoops!
    $endgroup$
    – RandomMathGuy
    Dec 8 '18 at 21:23











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

For $d$ to be a divisor of $n$, $d$ must be of the form $2^a cdot 3^b cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $0 leq a leq 14, 0 leq b leq 9, 0 leq c leq 8, 0 leq d leq 10, 0 leq eleq 3, 0leq f leq 5, 0 leq g leq 10$.



Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a cdot 3^color{red}{9} cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $color{red}{10} leq a leq 14, color{red}{2} leq c leq 8, color{red}{5} leq d leq 10, color{red}{2} leq eleq 3, color{red}{2}leq f leq 5, color{red}{2} leq g leq 10$.



Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c in {3,6}, d in {6,9}, e=3, f=3, g in {3,6,9}.$$
Thus the total number of choices we have are
$$2 cdot 2 cdot 3=12.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
    $endgroup$
    – DevAllanPer
    Dec 8 '18 at 21:34












  • $begingroup$
    @DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
    $endgroup$
    – Anurag A
    Dec 9 '18 at 0:45


















1












$begingroup$

For $d$ to be a divisor of $n$, $d$ must be of the form $2^a cdot 3^b cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $0 leq a leq 14, 0 leq b leq 9, 0 leq c leq 8, 0 leq d leq 10, 0 leq eleq 3, 0leq f leq 5, 0 leq g leq 10$.



Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a cdot 3^color{red}{9} cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $color{red}{10} leq a leq 14, color{red}{2} leq c leq 8, color{red}{5} leq d leq 10, color{red}{2} leq eleq 3, color{red}{2}leq f leq 5, color{red}{2} leq g leq 10$.



Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c in {3,6}, d in {6,9}, e=3, f=3, g in {3,6,9}.$$
Thus the total number of choices we have are
$$2 cdot 2 cdot 3=12.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
    $endgroup$
    – DevAllanPer
    Dec 8 '18 at 21:34












  • $begingroup$
    @DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
    $endgroup$
    – Anurag A
    Dec 9 '18 at 0:45
















1












1








1





$begingroup$

For $d$ to be a divisor of $n$, $d$ must be of the form $2^a cdot 3^b cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $0 leq a leq 14, 0 leq b leq 9, 0 leq c leq 8, 0 leq d leq 10, 0 leq eleq 3, 0leq f leq 5, 0 leq g leq 10$.



Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a cdot 3^color{red}{9} cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $color{red}{10} leq a leq 14, color{red}{2} leq c leq 8, color{red}{5} leq d leq 10, color{red}{2} leq eleq 3, color{red}{2}leq f leq 5, color{red}{2} leq g leq 10$.



Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c in {3,6}, d in {6,9}, e=3, f=3, g in {3,6,9}.$$
Thus the total number of choices we have are
$$2 cdot 2 cdot 3=12.$$






share|cite|improve this answer











$endgroup$



For $d$ to be a divisor of $n$, $d$ must be of the form $2^a cdot 3^b cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $0 leq a leq 14, 0 leq b leq 9, 0 leq c leq 8, 0 leq d leq 10, 0 leq eleq 3, 0leq f leq 5, 0 leq g leq 10$.



Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a cdot 3^color{red}{9} cdot 5^c cdot 7^d cdot 11^e cdot 13^f cdot 37^g$, where $color{red}{10} leq a leq 14, color{red}{2} leq c leq 8, color{red}{5} leq d leq 10, color{red}{2} leq eleq 3, color{red}{2}leq f leq 5, color{red}{2} leq g leq 10$.



Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c in {3,6}, d in {6,9}, e=3, f=3, g in {3,6,9}.$$
Thus the total number of choices we have are
$$2 cdot 2 cdot 3=12.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 21:30

























answered Dec 8 '18 at 21:24









Anurag AAnurag A

26.4k12251




26.4k12251












  • $begingroup$
    how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
    $endgroup$
    – DevAllanPer
    Dec 8 '18 at 21:34












  • $begingroup$
    @DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
    $endgroup$
    – Anurag A
    Dec 9 '18 at 0:45




















  • $begingroup$
    how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
    $endgroup$
    – DevAllanPer
    Dec 8 '18 at 21:34












  • $begingroup$
    @DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
    $endgroup$
    – Anurag A
    Dec 9 '18 at 0:45


















$begingroup$
how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
$endgroup$
– DevAllanPer
Dec 8 '18 at 21:34






$begingroup$
how would the answer change if it would have been 3^8 for both numbers instead of 3^9?
$endgroup$
– DevAllanPer
Dec 8 '18 at 21:34














$begingroup$
@DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
$endgroup$
– Anurag A
Dec 9 '18 at 0:45






$begingroup$
@DevAllanPer If it is $3^8$ for both $n$ and the number whose multiple we are looking for, then such a divisor cannot be a perfect cube.
$endgroup$
– Anurag A
Dec 9 '18 at 0:45













1












$begingroup$

Since we are required to be a multiple of $2^{10}cdot3^9cdot5^2cdot7^5cdot11^2cdot 13^2cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}cdot3^9cdot5^3cdot7^6cdot11^3cdot 13^3cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:20












  • $begingroup$
    @JMoravitz Whoops!
    $endgroup$
    – RandomMathGuy
    Dec 8 '18 at 21:23
















1












$begingroup$

Since we are required to be a multiple of $2^{10}cdot3^9cdot5^2cdot7^5cdot11^2cdot 13^2cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}cdot3^9cdot5^3cdot7^6cdot11^3cdot 13^3cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:20












  • $begingroup$
    @JMoravitz Whoops!
    $endgroup$
    – RandomMathGuy
    Dec 8 '18 at 21:23














1












1








1





$begingroup$

Since we are required to be a multiple of $2^{10}cdot3^9cdot5^2cdot7^5cdot11^2cdot 13^2cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}cdot3^9cdot5^3cdot7^6cdot11^3cdot 13^3cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.






share|cite|improve this answer











$endgroup$



Since we are required to be a multiple of $2^{10}cdot3^9cdot5^2cdot7^5cdot11^2cdot 13^2cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}cdot3^9cdot5^3cdot7^6cdot11^3cdot 13^3cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 21:23

























answered Dec 8 '18 at 21:19









RandomMathGuyRandomMathGuy

462




462








  • 1




    $begingroup$
    You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:20












  • $begingroup$
    @JMoravitz Whoops!
    $endgroup$
    – RandomMathGuy
    Dec 8 '18 at 21:23














  • 1




    $begingroup$
    You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
    $endgroup$
    – JMoravitz
    Dec 8 '18 at 21:20












  • $begingroup$
    @JMoravitz Whoops!
    $endgroup$
    – RandomMathGuy
    Dec 8 '18 at 21:23








1




1




$begingroup$
You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
$endgroup$
– JMoravitz
Dec 8 '18 at 21:20






$begingroup$
You seem to have missed the that are multiples of part of the problem statement. With $kleq 3$ you are not going to be a multiple of $2^{10}$ nor a multiple of any multiple of $2^{10}$
$endgroup$
– JMoravitz
Dec 8 '18 at 21:20














$begingroup$
@JMoravitz Whoops!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:23




$begingroup$
@JMoravitz Whoops!
$endgroup$
– RandomMathGuy
Dec 8 '18 at 21:23


















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