1968 AHSME Problems/Problem 21

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 $1! + 2! + 3! +  4! \equiv 3$ mod $10$, which means that the unit digit is $3$, we have our answer of $\fbox{D}$ as desired.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png