Given that $k = 21!/18!$ , which of the following options is the smallest that shares no prime factors with $k$ ?

Correct answer marked below

119

121

Correct

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Explanation

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Official Explanation

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To find

k

, we calculate

k = \frac{21!}{18!} = \frac{21 \times 20 \times 19 \times18!}{18!} = 21 \times 20 \times 19 = 3 \times 7 \times 2^2 \times 5 \times 19

. Now we analyze the options:

(A) 21 has a prime factor of 7, so it is OUT.

(B)

36 = 2^2 \times 3^2

, which shares prime factors 2 and 3 with k, so it is OUT.…

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What this question tests

Your ability to manipulate linear and quadratic expressions, substitute unknowns, and solve equations under time pressure.

Losing a sign when moving terms across the equals sign, or dividing both sides by a variable that could equal zero.