2000 AMC 10 Problems/Problem 11

Revision as of 15:06, 7 January 2009 by BOGTRO (talk | contribs) (New page: Two prime numbers between <math>4</math> and <math>18</math> are both odd. odd*odd=odd. odd-odd-odd=odd. Thus, we can discard the even choices. <math>ab-a-b=(a-1)(b-1)-1</math>. <mat...)
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Two prime numbers between $4$ and $18$ are both odd.

odd*odd=odd.

odd-odd-odd=odd.

Thus, we can discard the even choices.

$ab-a-b=(a-1)(b-1)-1$.

$(a-1)(b-1)-1=21,119,231$.

Both of these factors are even, so the number +1 must be a multiple of $4$.

$120$ is the only possiblity.

$11,13$ satisfy this, $11*13-11-13=143-24=119$.

C.

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