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

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==Problem==
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#REDIRECT [[2000 AMC 12 Problems/Problem 6]]
 
 
==Solution==
 
 
 
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>.
 
 
 
<math>(a-1)(b-1)-1=21,119,231</math>.
 
 
 
Both of these factors are even, so the number +1 must be a multiple of <math>4</math>.
 
 
 
<math>120</math> is the only possiblity.
 
 
 
<math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>.
 
 
 
C
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2000|num-b=10|num-a=12}}
 

Latest revision as of 22:42, 26 November 2011

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