Difference between revisions of "1996 AHSME Problems/Problem 14"
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+ | ==Problem== | ||
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+ | Let <math>E(n)</math> denote the sum of the even digits of <math>n</math>. For example, <math> E(5681) = 6+8 = 14 </math>. Find <math> E(1)+E(2)+E(3)+\cdots+E(100) </math> | ||
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+ | <math> \text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250 </math> | ||
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+ | ==Solution== | ||
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+ | The problem is asking for the sum of all the even digits in the numbers <math>1</math> to <math>100</math>. We can remove <math>100</math> from the list, add <math>00</math> to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list: | ||
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+ | <math>00, 01, 02, 03, ..., 10, 11, ..., 98, 99</math>. | ||
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+ | There are <math>2\cdot 100 = 200</math> digits on that list, and each digit appears <math>\frac{200}{10} = 20</math> times. | ||
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+ | Thus, each even digit appears <math>20</math> times, and the sum of all the even digits is <math>0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400</math>, and the correct answer is <math>\boxed{C}</math>. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=13|num-a=15}} | {{AHSME box|year=1996|num-b=13|num-a=15}} |
Revision as of 13:59, 19 August 2011
Problem
Let denote the sum of the even digits of . For example, . Find
Solution
The problem is asking for the sum of all the even digits in the numbers to . We can remove from the list, add to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list:
.
There are digits on that list, and each digit appears times.
Thus, each even digit appears times, and the sum of all the even digits is , and the correct answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |