The answer is?

Number Theory Level 4

Find the value of integer x x such that

1 ! 2 ! 3 ! 4 ! ⋯ 99 ! 100 ! x ! \dfrac{1! \ 2! \ 3! \ 4! \ \cdots \ 99! \ 100!}{x!}

is a perfect square.


The answer is 50.

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Keshav Bassi by Keshav Bassi , 20, India

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1 solution

Curtis Clement
Aug 24, 2015

∏ a = 1 100 a ! = 100 × 9 9 2 × 9 8 3 × . . . × 2 99 \prod_{a=1}^{100} a! = 100 \times\ 99^2 \times\ 98^3 \times\ ... \times\ 2^{99} Now taking out all the squares (1) (and even powers leaves us with: 98 × 96 × . . . × 2 = 2 49 49 ! \ 98 \times\ 96 \times\ ...\times\ 2 = 2^{49} 49! Now we want to divide this to leave a square number so we test x = 50 50 ! = 50.49 ! = 5 2 × 2.49 ! \ 50 ! = 50 . 49! = 5^2 \times 2 . 49! We ignore a factor of 25 as this is a square term that will cancel out from a term we neglected from (1). So we have: 2 49 . 49 ! 2 ( 49 ! ) = 4 24 \frac{2^{49} .49!}{2(49!)} = 4^{24} Hence, x = 50 \ x = 50

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. My approach was just a slight variation: We first observe that for ∏ n = 1 100 n ! \prod_{n=1}^{100} n! each odd n n appears an even number of times and each even n n appears an odd number of times. Thus we need to eliminate one of each of the even n n to end up with a perfect square. But

∏ n = 1 50 2 n = 2 50 ∗ ∏ n = 1 50 n = 2 50 ∗ 50 ! , \displaystyle\prod_{n=1}^{50} 2n = 2^{50} * \prod_{n=1}^{50} n = 2^{50}*50!,

so if we divide by 50 ! 50! we end up with a perfect square.

Brian Charlesworth - 5 years, 9 months ago

Nicely done Curtis :-)

Keshav Bassi - 5 years, 9 months ago

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