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| − | == Problem ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 11]] |
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| − | The positive integers <math>A</math>, <math>B</math>, <math>A-B</math>, and <math>A+B</math> are all prime numbers. The sum of these four primes is
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| − | <math> \mathrm{(A) \ } \text{even}\qquad \mathrm{(B) \ } \text{divisible by }3\qquad \mathrm{(C) \ } \text{divisible by }5\qquad \mathrm{(D) \ } \text{divisible by }7\qquad \mathrm{(E) \ } \text{prime}</math>
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| − | == Solution ==
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| − | The sum is <math>A+B+A-B+A+B=3A+B</math>. Since <math>A</math>, <math>A-B</math>, and <math>A+B</math> are all prime, they must all be odd, so <math>B=2</math>. A quick check gives <math>A=5</math>. Hence, the sum is <math>17</math>, which is prime. <math>\mathrm{ (E) \ }</math>
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