2007 iTest Problems/Problem 6

Revision as of 10:04, 16 December 2007 by 1=2 (talk | contribs) (New page: ==Problem== Find the units digit of the sum <cmath>\sum_{i=1}^{100}(i!)^{2}</cmath> <math>\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,3\quad\mathrm{(D)}\,5\quad\mathrm{(E)}\,7\...)
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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 $\Rightarrow \mathrm{(E)}$