2009 AMC 12B Problem 6

Attempt Problem 6 of the 2009 AMC 12B 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 2009 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

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?
22
33
44
55
66
Answer: C
Concepts:order of operationssystematic listing
Small hint:
The result depends on when the addition is performed relative to the two multiplications
Big hint:
Enumerate the distinct groupings; several parenthesizations collapse to the same value
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.
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