Perfect squares?

Number Theory Level 4

Let $N={ 1 }^{ 9 }\times { 2 }^{ 8 }\times { 3 }^{ 7 }\times { 4 }^{ 6 }\times { 5 }^{ 5 }\times { 6 }^{ 4 }\times { 7 }^{ 3 }\times { 8 }^{ 2 }\times { 9 }^{ 1 }$ . How many perfect squares divide $N$ ?

Source: old IWYMIC problem.

The answer is 672.

2 solutions

Danny He
Aug 1, 2014

Rewriting $N$ as a product of primes, we get that $N = 2^{30}*3^{13}*5^{5}*7^{3}$

Now since we want squares, we can tweak the divisor counting function such that it counts only square divisors.

First we note that the primes respectively have $16,7,3,$ and $2$ even powers.

So there are $16*7*3*2$ square factors of $N$ .

$16*7*3*2 = 672$

We could just make $2^{30}\times3^{13}\times5^{5}\times7^{3} = 4^{15}\times9^{6}\times25^{2}\times49^{1}\times3\times5\times7$

The number of perfect number factors = $16\times7\times3\times2 = 672$

Btw it's still the same though.

Samuraiwarm Tsunayoshi - 6 years, 10 months ago

i think n= 2^30 * 3^13 * 5^5 * 7^3 & i guess that doesnt make any difference

Alamuru Ganesh - 6 years, 10 months ago

Good problem @Isaac but I remember seeing the same problem with a different title on this site :)

Krishna Ar - 6 years, 10 months ago

Sorry, I didn't know the problem was already posted :(

Isaac Jiménez - 6 years, 10 months ago

It's ok...the problem was actually slightly different. Titled-perfect square divisors. You actully had to find the P.F of the factorials, but over here you've given them already. :D

Krishna Ar - 6 years, 10 months ago

Rahul Kumar
Oct 3, 2014

N=2^30 * 3^13 * 5^5 * 7^3 , :We know that for a number to be a perfect square, its factor must have the even number of powers. 2^(0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30) * 3^(0,2,4,6,8,10,12) * 5^(0,2,4) * 7^(0,2),Hence, the total number of factors which are perfect square are 16 7 3*2=672

Perfect squares?

Source: old IWYMIC problem.

The answer is 672.

2 solutions

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