2009 AMC 10A Problems/Problem 5: Difference between revisions
VelaDabant (talk | contribs) New page: ==Problem== What is the sum of the digits of the square of 111,111,111 ? <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</ma... |
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==Solution== | ==Solution== | ||
<math>\longrightarrow \fbox{E}</math> | Using the standard multiplication algorithm, <math>111111111^2=12345678987654321</math> whose digit sum is <math>81\longrightarrow \fbox{E}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | ||
Revision as of 18:50, 18 February 2009
Problem
What is the sum of the digits of the square of 111,111,111 ?
Solution
Using the standard multiplication algorithm, 
whose digit sum is 
See also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
