AMC 10 · 2013 · #3

Grade 4 counting
consecutive-integer-countsymmetry-argument work-backwardspattern-recognition ↑ Prerequisites: multi-digit-arithmetic
📏 Short solution 💡 2 insights
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Problem
A run of numbers is listed forwards and then backwards. Find one number's place in the backward list.

Pick an answer.

(A)
146
(B)
147
(C)
148
(D)
149
(E)
150
How to solve
Strategy Work Backwards

The question literally asks for a position in a list read from the end, so Tool #11 (Work Backwards) fits directly: start at 201 and count down until 53 is reached. Tool #7 (Identify Subproblems) supplies the one reusable fact this rests on — how many whole numbers lie in a stretch from one value to another, inclusive. Tool #5 (Look for a Pattern) gives a fast cross-check, since a forward position and a backward position in the same list always add to a fixed total.

1STEP 1

Count-inclusive rule

The forward list confirms the counting rule.

position (forward) = 53 - 3 + 1 = 51 ✓
2STEP 2

Reframe the backward count

The backward list counts from the other end.

n = 201 - 53 + 1
3STEP 3

Compute n

The same rule gives 149, choice (D).

n = 201 - 53 + 1 = 148 + 1 = 149 → (D)
Answer
149
The whole list holds 201 - 3 + 1 = 199 numbers. A number's forward position and backward position in the same list must add up to 199 + 1 = 200 (the +1 because the number itself is counted from both directions). Forward, 53 was the 51st, so backward it should be 200 - 51 = 149. This agrees with (D). It is also sensible that a number near the small end sits near the far end of the backward list, i.e. a large position like 149.
💡Key takeaway

To find a spot in a backward list, count how many numbers sit from the top end down to yours: 201 - 53 + 1 = 149, using include-both-ends counting you already know.

  • Count-inclusive rule
  • Reframe the backward count
  • Compute n