AMC 10 · 2021 · #3

Grade 6 number-theory
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
place-valuedigit-decompositionlinear-equations-one-varmulti-digit-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: place-value
📏 Short solution 💡 2 insights
Problem
Two natural numbers add to 17,402. The bigger one ends in zero, so it is a multiple of ten. Erasing that bigger number's units digit leaves the smaller number. Find the difference between the two numbers.

Pick an answer.

(A)
$10{,}272$
(B)
$11{,}700$
(C)
$13{,}362$
(D)
$14{,}238$
(E)
$15{,}426$
How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems) is the spine: name the smaller number s; then the bigger is 10s (since erasing the ones digit is the same as dividing by 10). The sum condition becomes one tiny equation 11s = 17,402, and the difference is 9s. Tool #3 (Eliminate) acts as a fast safety net — the difference ends in 9s, whose ones digit must be 8 (since s ends in 2). Only choice (D) 14,238 ends in 8.

1STEP 1

Both numbers, one letter

The bigger is ten times the smaller.

smaller = s, bigger = 10s
2STEP 2

Use the sum

The sum is eleven times it.

11s = 17,402 → s = 17,402/11 = 1,582
3STEP 3

Take the difference

Subtracting gives 14,238.

15,820 - 1,582 = 14,238 → (D)
4STEP 4

Check the last digit

The last digit alone narrows it to one choice.

ones digit of difference = 8 → (D)
Answer
14,238
Verify the trio: smaller + bigger = 1,582 + 15,820 = 17,402 ✓. Erasing the ones digit of 15,820 gives 1,582 ✓. Difference 14,238 is roughly 9/11 of 17,402 ≈ 14,238 ✓. The number lines up with choice (D), and no other choice has the right ones digit.
💡Key takeaway

This AMC 12 problem only needs Grade 6 "sliding digits left is multiplying by ten" — once you call the smaller number s, the bigger is 10s, and 11s = 17,402 gives s = 1,582, so the difference is 14,238.

More like this

Same archetype — closest grade level first.