1968 AHSME Problems/Problem 21

Revision as of 15:03, 11 February 2017 by Mathisfun04 (talk | contribs) (Solution)

Problem

If $S=1!+2!+3!+\cdots +99!$, then the units' digit in the value of S is:

$\text{(A) } 9\quad \text{(B) } 8\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 0$

Solution

Note that every factorial after $5!$ has a unit digit of $0$, meaning that we can disregard them. Thus, we only need to find the units digit of $1! + 2! + 3! +  4!$, and as $1! + 2! + 3! +  4! = 3 \equiv 10$, which means that the unit digit is $3$, we have our answer of $\fbox{D}$ as desired.

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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