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2005 AMC 10A Problems/Problem 16: Difference between revisions

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[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]
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Revision as of 10:30, 4 July 2013

Problem

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$. How many two-digit numbers have this property?

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19$

Solution

Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.

So $(10a+b)-(a+b)=9a$ must have a units digit of $6$

This is only possible if $9a=36$, so $a=4$ is the only way this can be true.

So the numbers that have this property are $40$, $41$, $42$, $43$, $44$, $45$, $46$, $47$, $48$, $49$.

Therefore the answer is $10\Rightarrow D$

See Also

These problems are copyrighted © by the Mathematical Association of America.

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