2007 iTest Problems/Problem 6

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Problem

Find the units digit of the sum

$\sum_{i=1}^{100}(i!)^{2}$

$\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,3\quad\mathrm{(D)}\,5\quad\mathrm{(E)}\,7\quad\mathrm{(F)}\,9$

Solution

If i is less than 5, then $i!$ has a positive units digit, if $i\geq 5$ , then $i!$ has a units digit of 0, as does $(i!)^2$ . So we only need to worry about i=1-4.

$(1!)^2=1$
$(2!)^2=4$
$(3!)^2=36$
$(4!)^2=576$
$1+4+6+6=17$ , which has a units digit of 7 $\Rightarrow \mathrm{(E)}$

2007 iTest (Problems)
Preceded by: Problem 5	Followed by: Problem 7
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2007 iTest Problems/Problem 6

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