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Ukrainian National Mathematical Olympiad

Algebra Difficulty 4.4 AIME Find the answer Ukraine

Find the exact value of the expression below:
(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)(6!−5!)(5!−4!)(4!−3!)(3!−2!)(2!−1)! \frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!}
if n!n! denotes the product 1⋅2⋅3⋯n1 \cdot 2 \cdot 3 \cdots n for every natural number nn.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Using the obvious equality (n+1)!=(n+1)⋅n!(n+1)! = (n+1) \cdot n! we can make the following transformations:
(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)(6!−5!)(5!−4!)(4!−3!)(3!−2!)(2!−1)!=5!(6+1)⋅4!(5+1)⋅3!(4+1)⋅2!(3+1)⋅1!(2+1)!5!(6−1)⋅4!(5−1)⋅3!(4−1)⋅2!(3−1)⋅1!(2−1)!=7⋅6⋅5⋅4⋅35⋅4⋅3⋅2⋅1=7⋅62⋅1=21. \frac{(6!+5!)(5!+4!)(4!+3!)(3!+2!)(2!+1!)}{(6!-5!)(5!-4!)(4!-3!)(3!-2!)(2!-1)!} = \frac{5!(6+1) \cdot 4! (5+1) \cdot 3! (4+1) \cdot 2! (3+1) \cdot 1! (2+1)!}{5!(6-1) \cdot 4! (5-1) \cdot 3! (4-1) \cdot 2! (3-1) \cdot 1! (2-1)!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6}{2 \cdot 1} = 21.

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