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2007 AMC 10A Problem 16

Problem 16 of 25IntermediateNumber TheoryCounting & Probability

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?

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Solution

Half the integers from 00 to 20072007 are odd, so each of adad and bcbc is odd with probability 1212=14\tfrac12 \cdot \tfrac12 = \tfrac14 and even with probability 34.\tfrac34. The difference adbcad - bc is even when both products have the same parity: 1414+3434=116+916=58. \tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58. Thus, the correct answer is E.

More practice

Concepts: parity · basic probability · independent events

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.