Difference between revisions of "2007 iTest Problems/Problem 6"
(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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<math>1+4+6+6=17</math>, which has a units digit of 7 <math>\Rightarrow \mathrm{(E)}</math> | <math>1+4+6+6=17</math>, which has a units digit of 7 <math>\Rightarrow \mathrm{(E)}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{iTest box|year=2007|num-b=5|num-a=7}} | ||
Revision as of 10:05, 16 December 2007
Problem
Find the units digit of the sum
![\[\sum_{i=1}^{100}(i!)^{2}\]](http://latex.artofproblemsolving.com/4/6/d/46d3c9f38511a89b99e238905fb6f3747dfd6176.png)
Solution
If i is less than 5, then 
has a positive units digit, if 
, then has a units digit of 0, as does 
. So we only need to worry about i=1-4.
, which has a units digit of 7 
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 5 |
Followed by: Problem 7 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 | ||
