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The factorial \((!)\) of a positive integer \(n\) denotes the product of all integers from \(1\) to \(n,\) inclusive. If

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The factorial \((!)\) of a positive integer \(n\) denotes the product of all integers from \(1\) to \(n,\) inclusive. If

by Gmat_mission » Thu Dec 17, 2020 1:10 pm

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The factorial \((!)\) of a positive integer \(n\) denotes the product of all integers from \(1\) to \(n,\) inclusive. If \(k = 1! + 2! + 3! + \cdots + p!,\) where \(p\) is a prime number greater than \(10,\) what is the remainder when \(k\) is divided by \(4?\)

A. 0

B. 1

C. 2

D. 3

E. 9

Answer: B

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Re: The factorial \((!)\) of a positive integer \(n\) denotes the product of all integers from \(1\) to \(n,\) inclusive

by Scott@TargetTestPrep » Wed Dec 23, 2020 4:05 pm
Gmat_mission wrote: ↑
Thu Dec 17, 2020 1:10 pm
 The factorial \((!)\) of a positive integer \(n\) denotes the product of all integers from \(1\) to \(n,\) inclusive. If \(k = 1! + 2! + 3! + \cdots + p!,\) where \(p\) is a prime number greater than \(10,\) what is the remainder when \(k\) is divided by \(4?\)

A. 0

B. 1

C. 2

D. 3

E. 9

Answer: B

Solution:

Recall that for any positive integers n and m where n ≥ m, n! is always divisible by m. That is because n! will always contain the factor of m when n ≥ m. Therefore, if p ≥ 4, p! will be divisible by 4. Thus, the remainder when k is divided by 4 depends only on the sum of the first 3 terms, which is 1! + 2! + 3! = 1 + 2 + 6 = 9. Since the remainder when 9 is divided by 4 is 1, the remainder when k is divided by 4 is also 1.

Answer: B

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