Find the remainder when (1!)^2 + (2!)^2 + (3! | vedclass

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(C) We need to find the remainder of $S = \sum_{n=1}^{100} (n!)^2$ when divided by $100$.Calculating the first few terms:$(1!)^2 = 1^2 = 1$$(2!)^2 = 2

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) We need to find the remainder of $S = \sum_{n=1}^{100} (n!)^2$ when divided by $100$.Calculating the first few terms:$(1!)^2 = 1^2 = 1$$(2!)^2 = 2^2 = 4$$(3!)^2 = 6^2 = 36$$(4!)^2 = 24^2 = 576 \equiv 76 \pmod{100}$$(5!)^2 = 120^2 = 14400 \equiv 0 \pmod{100}$For all $n \ge 5$,$n!$ contains at least two factors of $5$ and two factors of $2$,so $n!$ is a multiple of $10$. Thus,$(n!)^2$ is a multiple of $100$.Therefore,the sum $S \equiv 1 + 4 + 36 + 76 \pmod{100}$$S \equiv 117 \pmod{100}$$S \equiv 17 \pmod{100}$The remainder is $17$.

Find the remainder when $(1!)^2 + (2!)^2 + (3!)^2 + \dots + (100!)^2$ is divided by $10^2$.

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