AMC 10 · 2022 · #1

Grade 5 arithmetic
fraction-arithmeticfraction-multiplication identify-subproblemswork-backwards ↑ Prerequisites: fraction-arithmetic
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Problem
Evaluate the three-level continued fraction 3+13+13+133 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}} and match the value to one of the five answer choices.

Pick an answer.

(A)
$\frac{31}{10}$
(B)
$\frac{49}{15}$
(C)
$\frac{33}{10}$
(D)
$\frac{109}{33}$
(E)
$\frac{15}{4}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The expression has three nested layers, so we cannot evaluate it left-to-right. Tool #7 (Identify Subproblems) splits the giant fraction into three identical small jobs: "compute 3 + 1/(something already known)". Tool #11 (Work Backwards) tells us where to start — at the deepest layer, then climb outward, replacing each □ with the value we just found. Algebra (#13) is overkill since there are no unknowns; the layered structure tells us the path.

1STEP 1

Innermost layer: rewrite 3 as 93\frac{9}{3} and add to 13\frac{1}{3}, giving 103\frac{10}{3}.

3 + 13\frac{1}{3} = 93\frac{9}{3} + 13\frac{1}{3} = 103\frac{10}{3}
2STEP 2

Middle layer: dividing 1 by 103\frac{10}{3} flips it to 310\frac{3}{10}.

1103\frac{1}{\frac{10}{3}} = 1 ÷ 103\frac{10}{3} = 1 × 310\frac{3}{10} = 310\frac{3}{10}
3STEP 3

Finish the middle layer: 3 becomes 3010\frac{30}{10}, plus 310\frac{3}{10} gives 3310\frac{33}{10}.

3 + 310\frac{3}{10} = 3010\frac{30}{10} + 310\frac{3}{10} = 3310\frac{33}{10}
4STEP 4

Outer layer: flip 3310\frac{33}{10} to 1033\frac{10}{33}, add to 3 = 9933\frac{99}{33}, giving 10933\frac{109}{33}.

3 + 1033\frac{10}{33} = 9933\frac{99}{33} + 1033\frac{10}{33} = 10933\frac{109}{33} → (D)
Answer
10933\frac{109}{33}
Sanity check the size. Each layer is roughly 3 + 1/(a bit more than 3), so the whole thing should be a little more than 3 but well under 4. 10933\frac{109}{33} ≈ 3.303 fits that exactly. Choice (C) 3310\frac{33}{10} = 3.3 is the middle layer's value (a common trap if you stop one level too early); (D) is the genuine outer value, matching the answer.
💡Key takeaway

This AMC 10 problem only needs Grade 5 "add fractions with unlike denominators and flip a fraction" — work from the deepest 13\frac{1}{3} outward, one layer at a time.