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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 6]] |
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| − | Two different prime numbers between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
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| − | <math>\mathrm{(A)}\ 21 \qquad\mathrm{(B)}\ 60 \qquad\mathrm{(C)}\ 119 \qquad\mathrm{(D)}\ 180 \qquad\mathrm{(E)}\ 231</math>
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| − | ==Solution==
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| − | Two prime numbers between <math>4</math> and <math>18</math> are both odd.
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| − | odd*odd=odd.
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| − | odd-odd-odd=odd.
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| − | Thus, we can discard the even choices.
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| − | <math>ab-a-b=(a-1)(b-1)-1</math>.
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| − | <math>(a-1)(b-1)-1=21,119,231</math>.
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| − | Both of these factors are even, so the number +1 must be a multiple of <math>4</math>.
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| − | <math>120</math> is the only possiblity.
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| − | <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}}
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