2009 AMC 10B Problem 7

Attempt Problem 7 of the 2009 AMC 10B 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 10B solutions, or check the answer key.

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

7.

By inserting parentheses, it is possible to give the expression 2×3+4×52\times3+4\times5 several values. How many different values can be obtained?
22
33
44
55
66
Answer: C
Concepts:order of operationssystematic listing
Small hint:
Decide which operation is performed last
Big hint:
Try groupings such as (2×3)+(4×5)(2\times3)+(4\times5) and (2×3+4)×5(2\times3+4)\times5
Solution:
There are five full parenthesizations of four numbers joined by three operations. Evaluating all five gives
((2×3)+4)×5=50((2\times3)+4)\times5=50 and (2×(3+4))×5=70.(2\times(3+4))\times5=70.
(2×3)+(4×5)=26.(2\times3)+(4\times5)=26.
2×((3+4)×5)=702\times((3+4)\times5)=70 and 2×(3+(4×5))=46.2\times(3+(4\times5))=46.
Thus the distinct values are 26,46,50,26, 46, 50, and 70,70, for a total of 4.4.
Thus, the correct answer is C.
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