Question

What is the remainder when $2(8!)-21(6!)$ divides $14(7!)+14(13!)$?

• A. 1
• B. 7!
• C. 8!
• D. 9!

Answer 1

The correct option is B 7!

$\frac{[7!(14+14\times 13\times 12\times 11\times 10\times 9\times 8\left)\right]}{[7!(16-3\left)\right]}=\frac{\left[\right(14+14\times 13\times 12\times 11\times 10\times 9\times 8\left)\right]}{\left(13\right)}\to remainder1$
Hence, the original remainder must be 7! (because for the sake of simplification of the numbers in the question we have cut the 7! from the numerator and the denominator in the first step).