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 $E(n)$ denote the sum of the even digits of $n$. For example, $E(5681) = 6+8 = 14$. Find $E(1)+E(2)+E(3)+\cdots+E(100)$

$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$

Solution

The problem is asking for the sum of all the even digits in the numbers $1$ to $100$. We can remove $100$ from the list, add $00$ 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:

$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$.

There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.

Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$, and the correct answer is $\boxed{C}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AHSME Problems and Solutions
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