Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>\text | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>\text 111111111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math> | <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math> | ||
==Solution 1== | ==Solution 1== | ||
− | Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math> | + | Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>(E) 81.</math> |
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− | ==Solution 2 | + | ==Solution 2== |
We note that | We note that | ||
Line 19: | Line 18: | ||
and <math>1,111^2 = 1234321</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> | + | 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</math> | ||
+ | |||
+ | <math>=(1+2+3\cdots n)+(1+2+3+\cdots n-1)</math> | ||
+ | |||
+ | <math>=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}</math> | ||
+ | |||
+ | <math>=\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== | ||
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The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>. | The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | Note that any number when taken <math>\mod{9}</math> yields the digit sum of that number. So, the problem has simplified to finding <math>111,111,111^2 \pmod{9}</math>. We note that <math>111,111,111 \mod{9}</math> is <math>9</math>, so <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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:45, 21 July 2021
Problem
What is the sum of the digits of the square of ?
Solution 1
Using the standard multiplication algorithm, whose digit sum is
Solution 2
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 .
Solution 4
Note that any number when taken yields the digit sum of that number. So, the problem has simplified to finding . We note that is , so .
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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