Difference between revisions of "2000 AMC 10 Problems/Problem 11"

(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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==Problem==
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==Solution==
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Two prime numbers between <math>4</math> and <math>18</math> are both odd.
 
Two prime numbers between <math>4</math> and <math>18</math> are both odd.
  
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<math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>.
 
<math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>.
  
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C
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==See Also==
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{{AMC10 box|year=2000|num-b=10|num-a=12}}

Revision as of 19:40, 8 January 2009

Problem

Solution

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

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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