Equating consecutive product

Number Theory Level 3

We know that $1\times 2 \times 3 \times \cdots \times 7 = 7\times \cdots \times 10.$ Now, for $n >7,$ can the following equality ever hold true: $1\times 2 \times 3 \times \cdots \times (n-1) \times n = n \times \cdots\times (n+k)$ for some positive integer $k?$

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1 solution

Vimay MarCisse
Jul 11, 2018

7! = 2 × 3 × 4 × 5 × 6 × 7

= 2 × 3 × 2 × 2 × 5 × 2 × 3 × 7 

= 2^4 × 3^2 × 5 × 7

= 7 × 2^3 × 3^2 × 2 × 5

= 7 × 8 × 9 × 10

Here i just factorised with prime factors. Now let's take 8!. Obviously, 8 × 9 × 10 is not equal to 8! (It is not enought). Thus we have to add "× 11". But 11 is a prime number. That's means that it can not be the result of "a × b" where a and b are both integers. Thus, the statement is wrong.

Equating consecutive product

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