How many divisors of 72^72 are perfect cubes?

Question

How many divisors of 72^72 are perfect cubes?
a) 3672
b) 3577
c) 2812
d) 3600
e) 7200

Supermaverick · Accepted Answer

Thanks IanStewart for the solution.I have found an alternate solution to solve it fast:

72^72 = (2^3 * 3^2 )^72 = (2^72)^3 * (3^48)^3
as both (2^72)^3 and (3^48)^3 are perfect cubes the total number of factor using factor property would be (72+1)*(48+1)=73 *49 = 3577

hellosanthosh2k2 · Answer

Prime factorization of  72^{72}  =  (2^3 * 3^2)^{72} 
 =  ((2^3)^{72} * 3^{144}) 
to find factors which are perfect cubes, we represent it as =  8^{72} * (3^3)^{48}  =  8^{72} * 27^{48} 
 since 8, 27 are perfect cubes, any combination powers of 8,27 must be perfect cube
 A: {  8^0 ,  8^1 ,  8^2  ,........................... 8^{72} } = total = 73
 B: {  27^0 ,  27^1 , ............................... 27^{48} } = total = 49
 number of factors which are perfect cubes:  73 * 49  =  3577  (B)

IanStewart · Answer

The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 * 3^6 is a perfect cube (it is the cube of 2^5 * 3^2), while 2^8 * 3^7 is not

If we prime factorize 72^72, we get (2^3 * 3^2 )^72 = 2^216 * 3^144

For a divisor of this to be a perfect cube, it needs to look like this:

2^a * 3^b

where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:

0, 3, 6, 9, ..., 213, 216

and b must be in this list:

0, 3, 6, ...., 141, 144

There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 73*49 choices in total for a and b together, and since the units digit of 73*49 is 7, the only possible answer among the choices is 3577.

Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.

chetan2u · Answer

such Qs when we are looking for ODD factors or EVEN factors or the one asked here, following method is best..

Number of factors of a number   x=a^p*b^r*c^s...  is ( p+1)(q+1)(s+1)...

here we are looking for CUBES..
so  72^{72}  can be written as  (2^3*3^2)^{24*3} = (2^{3*24}*3^{2*24})^3 = (2^{72}*3^{48})^3 
so cubes will be  (72+1)(48+1) = 73*49 = 3577 
B

versewriter · Answer

Here is the solution:

The question is asking for factors of a number which is a perfect cube in the given equation i.e. 72^72. So simplify the equation.

72^72 
= (3*3*8)^72
= (3^3*2^3)^72
= (3^144 * 2^216)
= 27^48 * 8^72
so the prime factors are (48+1)*(72+1) = 3577.

So the Answer is Option B

Adit_ · Answer

This is just twisted in a way to eventually find the total divisors.

72^72 = 2^(3)*72 * 3^(2)*72

= 2^216 * 3^144
= 2^(72*3) * 3^(48*3)

Total factors is simply (x+1)(y+1) 
= 73*49 = 3577.

Answer:  Option B

HubristicPsyche · Answer

Can someone help me understand the factor property here?

Thanks!

Adit_ · Answer

When u prime factorize a number, no.of factors of that number = Product of the powers + 1

Like 12 = 2^2*3 
Total positive factors of 12 = (2+1)(1+1)  
= 3*2 = 6
So 12 has 6 factors in total.

How many divisors of 72^72 are perfect cubes?

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How many divisors of 72^72 are perfect cubes?

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