Competition · AMC preparation · step 4 of 4

AMC 8 · 2025 · #16

Grade 4 countingarithmetic
Archetype Pair-Sum Invariant └ Sub-archetype Forbidden-Complement Selection
pattern-recognitionset-partitionsequences-arithmeticcombinations-basic pattern-recognitionidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticset-partition
📏 Medium solution 💡 3 insights
📘 View easy version →
Problem
From the numbers 1 through 10, pick any 5 distinct numbers. From 11 through 20, pick any 5 more distinct numbers. The catch: no chosen pair may differ by exactly 10 (so you cannot choose both 3 and 13, for instance). What is the sum of all ten chosen numbers?

Pick an answer.

(A)
95
(B)
100
(C)
105
(D)
110
(E)
115

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Organize Information in More Ways

The twenty numbers 1 through 20 are naturally listed in a straight row, but that layout hides the structure. Tool #15 says: re-organize. Pair each low number with its 'twin' that is 10 larger — (1, 11), (2, 12), …, (10, 20) — giving 10 pairs. The rule 'no two chosen numbers differ by exactly 10' becomes the much simpler rule 'do not pick both members of the same pair'. Tool #9 (Easier Problem) sanity-checks the idea on a tiny version, and tool #7 (Subproblems) lets us compute the total as (sum of the 10 low twins) + (extra 10 for every pair whose HIGH twin was chosen).

1STEP 1

Pair the numbers ten apart

Regroup the twenty numbers into ten twin-pairs differing by 10, so the rule just means 'never take both in one pair'.

Pairs: (k, k+10) for k = 1, 2, …, 10
2STEP 2

Count one pick per pair

You make 10 picks and there are exactly 10 pairs, so with no pair giving two you take exactly one from every pair.

10 picks ÷ 10 pairs = 1 pick per pair
3STEP 3

Test a smaller version

Sanity-check a tiny version with 4 pairs: every legal choice sums to 18, confirming the total can't change.

Small-case formula: (1+2+3+4) + 4 × 2 = 10 + 8 = 18 ✓
4STEP 4

Return to the full problem

Take one from each pair; picking a high twin adds 10, and exactly 5 highs are chosen, so the bonus above all-lows is 10 × 5 = 50.

Total = (1+2+…+10)_if we picked all lows + 10 × 5
5STEP 5

Add the low numbers and bonus

Add the low twins 1+2+…+10 = 55 to the 50 bonus to get 105.

Total = 55 + 50 = 105 → (C)
Answer
105
The total of ALL twenty numbers is 1 + 2 + … + 20 = 210. We are picking exactly half of them (10 of 20), so a reasonable answer should sit somewhere near half of 210, which is 105. Our answer 105 matches exactly — and the elegant reason is that we are forced to take exactly one number from each (low, high) pair, splitting the total 210 into two equal halves of 105 no matter which side of each pair we pick. The other choices (95, 100, 110, 115) would each require either picking more lows than highs or vice versa, which the 5-and-5 split forbids.
💡Key takeaway

This AMC 8 problem only needs Grade 4 number patterns you already know — once you pair each small number with its 'twin' that is 10 bigger, the answer pops out!

  • Pair the numbers ten apart
  • Count one pick per pair
  • Test a smaller version
  • Return to the full problem
  • Add the low numbers and bonus

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