How many perfect squares are divisors of the product
N = 1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! =
2*
2*3*
2*3*4*
2*3*4*5*
2*3*4*5*6*
2*3*4*5*6*7*
2*3*4*5*6*7*8*
2*3*4*5*6*7*8*9
looking down those vertical columns that's equal to:
(2^8)(3^7)(4^6)(5^5)(6^4)(7^3)(8^2)(9^1)
Break the composites into primes:
(2^8)(3^7)((2^2)^6)(5^5)((2*3)^4)(7^3)((2^3)^2)(3^2) =
(2^8)(3^7)(2^12)(5^5)(2^4*3^4)(7^3)(2^6)(3^2) =
(2^30)(3^13)(5^5)(7^3)
Every divisor of N is of the form (2^p)(3^q)(5^r)(7^s), where
0 <= p <= 30
0 <= q <= 13
0 <= r <= 5
0 <= s <= 3
The divisors which are perfect squares have even exponents,
(including 0).
0 <= p <= 30 contains 16 even exponents and 15 odd exponents
0 <= q <= 13 contains 7 even exponents and 7 odd exponents
0 <= r <= 5 contains 3 even exponents and 3 odd exponents
0 <= s <= 3 contains 2 even exponents and 2 odd exponents
Therefore for divisor (2^p)(3^q)(5^r)(7^s),
there are 16 choices for p, 7 choices for q, 3 choices for r,
and 2 choices for s.
Answer: 16*7*3*2 = 672 perfect square divisors of N.
Edwin
