AMC 10 · 2009 · #6

Grade 5 algebra
order-of-operationssystematic-enumeration systematic-enumeration ↑ Prerequisites: order-of-operations
📏 Medium solution 💡 2 insights
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Problem
Parentheses may be inserted into a fixed expression to force some operations first. Count how many different values it can equal.

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
How to solve
Strategy Make a Systematic List

The question is 'how many different values', so the safe route is to list every legal grouping and evaluate it (Tool #2). Four numbers in a row have exactly five full parenthesizations, a small enough set to write out completely with no guessing. Each grouping is then a short arithmetic subproblem to evaluate (Tool #7). Finally, because two groupings can land on the same number, Tool #3 (Eliminate Possibilities) is used to drop the duplicates so only distinct values remain.

1STEP 1

See what parentheses control

Parentheses only change the order of the operations.

2×3+4×5 → 5 possible groupings
2STEP 2

Write out the five groupings

There are exactly five legal groupings.

((2×3)+4)×5, (2×(3+4))×5, (2×3)+(4×5), 2×((3+4)×5), 2×(3+(4×5))
3STEP 3

Evaluate each grouping

Evaluating each gives its own number.

50, 70, 26, 70, 46
4STEP 4

Count the distinct values

Two of them repeat, so the count is 4, choice (C).

{50,70,26,70,46}→{26,46,50,70}→ 4 (C)
Answer
4
The two multiplications 2×3 and 4×5 can each be forced or delayed by grouping, and the smallest possible result comes from doing both multiplications first, 26, while the largest comes from folding the + inside so everything gets multiplied, 70. The four surviving values 26,46,50,70 all sit inside that range and are clearly different, so 4 is believable and matches choice (C).
💡Key takeaway

Parentheses only change which operation goes first, so list every grouping, work each one out, and count the answers that are actually different.

  • See what parentheses control
  • Write out the five groupings
  • Evaluate each grouping
  • Count the distinct values