Difference between revisions of "2009 AMC 10A Problems/Problem 5"
(→Solution 4) |
(→Solution 2 -- Find and Harness a Pattern) |
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(I hope you didn't seriously multiply it outright...) | (I hope you didn't seriously multiply it outright...) | ||
− | ==Solution 2== | + | ==Solution 2 -- Find And Harness a Pattern== |
− | + | We note that | |
− | <math> | + | <math>1^2 = 1</math>, |
− | We | + | <math>11^2 = 121</math>, |
+ | |||
+ | <math>111^2 = 12321</math>, | ||
+ | |||
+ | and <math>1,111^2 = 1234321</math>. | ||
+ | |||
+ | We can clearly see the pattern: If <math>X</math> is <math>111\cdots111</math>, with <math>n</math> ones (and for the sake of simplicity, assume that <math>n<10</math>), then the sum of the digits of <math>X^2</math> is <math>1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1=(1+2+3\cdots n)+(1+2+3+\cdots n-1)=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2</math>. Aha! We know that <math>111,111,111</math> has <math>9</math> digits, so its digit sum is <math>9^2=\boxed{81(E)}</math>. | ||
==Solution 3== | ==Solution 3== |
Revision as of 12:32, 30 December 2018
Problem
What is the sum of the digits of the square of ?
Solution 1
Using the standard multiplication algorithm, whose digit sum is (I hope you didn't seriously multiply it outright...)
Solution 2 -- Find And Harness a Pattern
We note that
,
,
,
and .
We can clearly see the pattern: If is , with ones (and for the sake of simplicity, assume that ), then the sum of the digits of is . Aha! We know that has digits, so its digit sum is .
Solution 3
We see that can be written as .
We can apply this strategy to find , as seen below.
The digit sum is 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 |
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