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2009 AMC 12B Problem 6

Problem 6 of 25EasierAlgebraCounting & Probability

By inserting parentheses, it is possible to give the expression 2×3+4×52 \times 3 + 4 \times 5 several values. How many different values can be obtained?

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Solution

The genuinely different groupings give (2×3)+(4×5)=26,(2 \times 3) + (4 \times 5) = 26,  ((2×3)+4)×5=50,\ ((2 \times 3) + 4) \times 5 = 50,  2×(3+(4×5))=46,\ 2 \times (3 + (4 \times 5)) = 46, and  2×(3+4)×5=70.\ 2 \times (3 + 4) \times 5 = 70. These are all distinct, so 44 values can be obtained. Thus, the correct answer is C.

More practice

Concepts: order of operations · systematic listing

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