Remainder infinity

Number Theory Level 3

Find the remainder of the following expression:

$\dfrac{1!^2 + 2!^2 + 3!^2 + \cdots + 100!^2}{36 }$

0 1 5 Too long to solve

2 solutions

Solomon Olayta
Dec 15, 2015

Since $36=2^{2} 3^{2}=6^{2}=(3!)^{2}$ , then, for each $k>=3$ , $k!^{2}$ is divisible by 36. Hence, ${3!^2 + \cdots + 100!^2}$ is divisible by 36. Thus, ${1!^2 + 2!^2 + 3!^2 + \cdots + 100!^2}=1!^2+2!^2+36k=5+36k$ . Therefore the remainder is 5.

Same approach, nice work.

tanay gaurav - 5 years, 6 months ago

Tanay Gaurav
Dec 12, 2015

We know 3! = 6 and squaring it will give 36. So, remainder of the whole expression is (1!^2 + 2!^2)/36 = 5 remainder(answer)

Remainder infinity

2 solutions

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