A number theory problem by Achal Jain

Number Theory Level 3

x @ y = x ( x + 1 ) ( x + 2 ) ⋯ ( x + y − 1 ) \large x@y = x (x+1)(x+2) \cdots (x+y - 1)

The operator @ @ is defined for positive integers x x and y y as above. Let h h denote the greatest common divisor of x @ y x @y and y ! y! . Which of the following options must be true?

Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

h = x y h=xy h > y ! h>y! h = y ! h=y! h = x ! h=x!

Achal Jain by Achal Jain , 19, India

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1 solution

x @ y = x ( x + 1 ) ( x + 2 ) ⋯ ( x + y − 1 ) = ( x + y − 1 ) ! ( x − 1 ) ! = ( x + y − 1 ) ! ( x − 1 ) ! y ! ⋅ y ! = ( x + y − 1 y ) y ! \begin{aligned} x@y & = x(x+1)(x+2)\cdots (x+y-1) \\ & = \frac {(x+y-1)!}{(x-1)!} \\ & = \frac {(x+y-1)!}{(x-1)!y!} \cdot y! \\ & = {x+y-1 \choose y }y! \end{aligned}

Note that ( x + y − 1 y ) {x+y-1 \choose y } is a positive integer, therefore x @ y x@y is a multiple of y ! y! and gcd ⁡ ( x @ y , y ! ) = y ! \gcd (x@y, y!) = y! , ⟹ h = y ! \implies \boxed{h = y!} .

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