Competition · AMC preparation · step 4 of 4

AMC 8 · 2003 · #4

Grade 4 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
systems-of-equationslinear-equations-two-varsystematic-enumeration convert-to-algebrasystematic-enumerationguess-and-check ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var
📏 Short solution 💡 2 insights
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Problem
A group of 7 children rode past on bicycles and tricycles, and Billy Bob counted 19 wheels total. How many of the children were riding tricycles?

Pick an answer.

(A)
2
(B)
4
(C)
5
(D)
6
(E)
7

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

There are only 8 possible tricycle counts (0 through 7), so testing values is faster than algebra. Tool #6 (Guess and Check) makes that direct: pick a tricycle count, pair it with the matching bicycle count, and add the wheels. Tool #5 (Look for a Pattern) makes the search even shorter — every time we swap one bicycle for one tricycle, the total wheel count goes up by exactly 1. That "+1 wheel per swap" rule pins down the answer in a single jump instead of trying every option.

1STEP 1

Start with all bicycles

Take all bicycles as the baseline: if all 7 children rode bicycles, the wheels would total 7 × 2 = 14.

7 × 2 = 14 wheels
2STEP 2

Find the missing wheels

The target is 19 wheels and the baseline gives 14, so we are short by 19 - 14 = 5 wheels.

19 - 14 = 5 extra wheels needed
3STEP 3

Find the gain per swap

Swapping one bicycle (2 wheels) for one tricycle (3 wheels) raises the total by 3 - 2 = 1 wheel per swap.

3 - 2 = 1 wheel per swap
4STEP 4

Swap five bicycles for tricycles

Closing the gap of 5 wheels takes 5 swaps, so there are 5 tricycles and 7 - 5 = 2 bicycles.

tricycles = 5, bicycles = 2 → (C)
Answer
5
Check the totals directly. With 5 tricycles and 2 bicycles: children = 5 + 2 = 7 (matches), and wheels = 5 × 3 + 2 × 2 = 15 + 4 = 19 (matches). The answer also has to be between the all-bike total of 14 and the all-trike total of 21, and 19 falls inside that range — so a valid mix must exist. The trap choices 4, 6, and 7 each fail the wheel count: 4 trikes give 4 × 3 + 3 × 2 = 18, 6 trikes give 6 × 3 + 1 × 2 = 20, and 7 trikes give 21.
💡Key takeaway

Start with the simplest guess (all bicycles), then notice that each bike-to-trike swap adds exactly one wheel — the missing wheel count tells you the number of tricycles directly. This AMC 8 problem becomes a Grade 4 multistep word problem, no algebra required.

  • Start with all bicycles
  • Find the missing wheels
  • Find the gain per swap
  • Swap five bicycles for tricycles

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