Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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Using the standard multiplication algorithm, <math>111111111^2=12345678987654321</math> whose digit sum is <math>81\longrightarrow \fbox{E}</math> | Using the standard multiplication algorithm, <math>111111111^2=12345678987654321</math> whose digit sum is <math>81\longrightarrow \fbox{E}</math> | ||
+ | Or | ||
+ | |||
+ | Add up all the ones(thus deriving the sum of the number) of <math>111,111,111</math> gives us <math>1+1+1+\dots+1 = 9</math> Thus, <math>(9)^2 = 81</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 09:29, 21 March 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
Or
Add up all the ones(thus deriving the sum of the number) of gives us Thus,
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 |