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AMC 8 · 2025 · #16

Grade 4 countingarithmetic
Archetype Pair-Sum Invariant
pattern-recognitionset-partitionsequences-arithmeticcombinations-basic pattern-recognitionidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticset-partition
📏 Medium solution 💡 3 insights
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Problem

Five distinct integers from 11 to 1010 are chosen, and five distinct integers from 1111 to 2020 are chosen. No two numbers differ by exactly 1010. What is the sum of the ten chosen numbers?

Pick an answer.

(A)
95
(B)
100
(C)
105
(D)
110
(E)
115
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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Low set: $5$ distinct integers chosen from $\{1, 2, \dots, 10\}$; High set: $5$ distinct integers chosen from $\{11, 12, \dots, 20\}$; No two chosen numbers differ by exactly $10$; Answer choices: (A) $95$, (B) $100$, (C) $105$, (D) $110$, (E) $115$

Unknowns: The sum of the $10$ chosen numbers

Understand

Restated: 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?

Givens: Low set: $5$ distinct integers chosen from $\{1, 2, \dots, 10\}$; High set: $5$ distinct integers chosen from $\{11, 12, \dots, 20\}$; No two chosen numbers differ by exactly $10$; Answer choices: (A) $95$, (B) $100$, (C) $105$, (D) $110$, (E) $115$

Plan

Primary tool: #15 Organize Information in More Ways

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

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 — $(1, 11), (2, 12), \dots, (10, 20)$ — 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).

Execute — Answer: C

#15 Organize Information in More Ways 4.OA.C.5 Step 1
  • Re-arrange the $20$ numbers into $10$ vertical pairs, each pair differing by exactly $10$: $(1,11), (2,12), (3,13), (4,14), (5,15), (6,16), (7,17), (8,18), (9,19), (10,20)$.
  • The forbidden 'differ by $10$' rule now means 'never pick both numbers in the same column'.
$$\text{Pairs: } (k,\ k+10)\ \text{ for }\ k = 1, 2, \dots, 10$$

💡 Re-arranging a list into a pattern (here, a $2 \times 10$ grid of paired numbers) is the Grade 4 idea of generating and using a number pattern.

#7 Identify Subproblems 4.OA.A.3 Step 2
  • Count: we pick $5$ low numbers and $5$ high numbers, $10$ picks total.
  • There are exactly $10$ pairs.
  • Since we cannot pick both members of a pair, each pick must come from a different pair — so we pick exactly ONE number from EACH of the $10$ pairs.
  • (If even one pair were skipped, two of our picks would have to share another pair, breaking the rule.)
$$10 \text{ picks} \div 10 \text{ pairs} = 1 \text{ pick per pair}$$

💡 Splitting a $10$-pick problem into $10$ tiny 'pick one of two' subproblems is the Grade 4 multi-step word-problem move.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 3
  • Sanity-check on a tiny version: lows $\{1, 2, 3, 4\}$, highs $\{5, 6, 7, 8\}$ (twins differ by $4$); pick $2$ from each side with no pair differing by exactly $4$.
  • Forbidden pairs: $(1,5), (2,6), (3,7), (4,8)$.
  • Try lows $\{1, 2\}$ and highs $\{7, 8\}$: sum $= 1 + 2 + 7 + 8 = 18$.
  • Try lows $\{3, 4\}$ and highs $\{5, 6\}$: sum $= 3 + 4 + 5 + 6 = 18$.
  • Try lows $\{1, 3\}$ and highs $\{6, 8\}$: sum $= 1 + 3 + 6 + 8 = 18$.
  • The total is invariant — exactly what we conjectured.
$$\text{Small-case formula: } (1+2+3+4) + 4 \times 2 = 10 + 8 = 18 \quad\checkmark$$

💡 Trying the same problem with tiny numbers ($4$ pairs instead of $10$) is the Grade 4 'solve a simpler related word problem first' habit.

#7 Identify Subproblems 4.OA.A.3 Step 4
  • Back to the original.
  • From each pair $(k, k+10)$ we pick exactly one number — either $k$ or $k+10$.
  • Notice: if we picked the high twin instead of the low twin, the contribution went UP by exactly $10$.
  • So $\text{Total} = (\text{sum of all low twins}) + 10 \times (\text{how many high twins we chose})$.
  • We know exactly $5$ of our $10$ picks came from the high side, so the second term is $10 \times 5 = 50$.
$$\text{Total} = \underbrace{(1+2+\dots+10)}_{\text{if we picked all lows}} \;+\; 10 \times 5$$

💡 Computing 'baseline plus bonus' separately is the Grade 4 multi-step word-problem strategy.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Add up the low twins using the well-known sum $1 + 2 + \dots + 10 = 55$, then add the $50$ bonus.

$$\text{Total} = 55 + 50 = 105 \;\Rightarrow\; \textbf{(C)}$$

💡 Fluent addition of multi-digit whole numbers ($55 + 50 = 105$) is the core Grade 4 NBT skill.

[1] #15 4.OA.C.5 Re-arrange the $20$ numbers into $10$ vertical pairs, each pair differing by exa
[2] #7 4.OA.A.3 Count: we pick $5$ low numbers and $5$ high numbers, $10$ picks total. There are
[3] #9 4.OA.A.3 Sanity-check on a tiny version: lows $\{1, 2, 3, 4\}$, highs $\{5, 6, 7, 8\}$ (t
[4] #7 4.OA.A.3 Back to the original. From each pair $(k, k+10)$ we pick exactly one number — ei
[5] #7 4.NBT.B.4 Add up the low twins using the well-known sum $1 + 2 + \dots + 10 = 55$, then ad

Review

Reasonableness: The total of ALL twenty numbers is $1 + 2 + \dots + 20 = 210$. 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.

Alternative: Tool #6 (Guess & Check) gives a fast confirmation: pick any valid example, like lows $\{1, 2, 3, 4, 5\}$ and highs $\{16, 17, 18, 19, 20\}$ (none differ by $10$). Sum $= (1+2+3+4+5) + (16+17+18+19+20) = 15 + 90 = 105$. Try another, lows $\{1, 3, 5, 7, 9\}$ and highs $\{12, 14, 16, 18, 20\}$: sum $= 25 + 80 = 105$. Same answer — confirming the invariant we proved with tool #15 + #7.

CCSS standards used (min grade 4)

  • 4.OA.C.5 Generate a number or shape pattern following a given rule (Reorganizing the $20$ numbers into the $10$ pairs $(k, k+10)$ so the 'differ by $10$' rule becomes 'pick one per pair'.)
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Reasoning that $10$ picks across $10$ pairs forces exactly one pick per pair, then computing $\text{total} = 55 + 10 \times 5$.)
  • 4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $55 + 50 = 105$ to obtain the final answer, and computing $1 + 2 + \dots + 10 = 55$.)

⭐ 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!

⭐ 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!

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