2007 AMC 12A Problem 12

Attempt Problem 12 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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12.

Integers a,a, b,b, c,c, and d,d, not necessarily distinct, are chosen independently and at random from 00 to 2007,2007, inclusive. What is the probability that adbcad-bc is even?

38\dfrac{3}{8}

716\dfrac{7}{16}

12\dfrac{1}{2}

916\dfrac{9}{16}

58\dfrac{5}{8}

Answer: E
Concepts:paritybasic probability
Difficulty rating: 1410
Solution:

Exactly half of the integers from 00 to 20072007 are odd.

A product adad is odd only when both factors are odd, with probability 1212=14,\tfrac12\cdot\tfrac12=\tfrac14, and even with probability 34.\tfrac34. The same holds for bc.bc.

Then adbcad-bc is even when both products are odd or both are even: 1414+3434=1016=58.\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58.

Thus, the correct answer is E.

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