AMC 10 · 2012 · #9
Grade 2 arithmeticTwo integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Six integers are added two at a time. The first two sum to $26$. Adding the next two makes the running total $41$. Adding the last two makes the total $57$. Find the fewest of the six integers that can be odd.
Givens: First pair sums to $26$; After adding the second pair, the running total is $41$; After adding the third pair, the running total is $57$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The minimum possible number of odd integers among the six
Understand
Restated: Six integers are added two at a time. The first two sum to $26$. Adding the next two makes the running total $41$. Adding the last two makes the total $57$. Find the fewest of the six integers that can be odd.
Givens: First pair sums to $26$; After adding the second pair, the running total is $41$; After adding the third pair, the running total is $57$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #11 Work Backwards
The question asks for a minimum, which is the direct signal for Tool #14 (Extreme Principle): push the six numbers to the boundary where odds are as scarce as possible. Tool #11 (Work Backwards) recovers each pair's own sum from the three running totals by subtraction. Tool #7 (Identify Subproblems) then treats the three pairs independently, because each pair's sum by itself decides the fewest odds that pair can hold.
Execute — Answer: A
2.NBT.B.5 Step 1 Recover each pair's sum
- The integers arrive in three pairs.
- Recover each pair's own contribution by subtracting consecutive running totals: the first pair is $26$, the second pair adds $41-26$, and the third pair adds $57-41$.
💡 A running total minus the previous running total is exactly what the newest pair added.
2.OA.C.3 Step 2 Parity of a pair sum
- Look at each pair on its own and ask about parity.
- Two integers add to an even number when they are both even or both odd (that is $0$ or $2$ odds); they add to an odd number only when exactly one of the two is odd.
💡 An odd number only survives in a sum when it is unpaired — two odds cancel back to even.
2.OA.C.3 Step 3 Push each pair to fewest odds
- Apply the extreme principle: give every pair the fewest odds its sum allows.
- Pair sums $26$ and $16$ are even, so each can be built from two even numbers — $0$ odds.
- Pair sum $15$ is odd, so it is forced to contain exactly $1$ odd.
💡 You can zero out the even-sum pairs, but nothing can drive an odd-sum pair below one odd.
2.NBT.B.5 Step 4 Add the minimums
- Add the per-pair minimums.
- A concrete build that hits every running total using just one odd is $(12, 14)$, $(7, 8)$, $(8, 8)$: the totals climb $26$, then $41$, then $57$, and $7$ is the only odd number.
- So the minimum is $0 + 1 + 0 = 1$, choice $\textbf{(A)}$.
💡 Three independent minimums simply add, and the odd-sum pair is the only one that must donate an odd.
2.NBT.B.5 The integers arrive in three pairs. Recover each pair's own contribution by subt 2.OA.C.3 Look at each pair on its own and ask about parity. Two integers add to an even n 2.OA.C.3 Apply the extreme principle: give every pair the fewest odds its sum allows. Pai 2.NBT.B.5 Add the per-pair minimums. A concrete build that hits every running total using Review
Reasonableness: Only the pair whose sum is odd ($15$) is forced to hold an odd number, and one odd there is unavoidable because two evens or two odds always make an even sum. The other two pairs (sums $26$ and $16$) are cleanly all-even, contributing zero. The explicit build $(12, 14)$, $(7, 8)$, $(8, 8)$ realizes exactly one odd while matching the totals $26, 41, 57$, so $1$ is achievable and cannot be beaten — answer (A).
Alternative: Tool #16 (Change Focus / Count the Complement) using mod $2$: replace each integer by its remainder mod $2$. The running totals $26, 41, 57$ are even, odd, odd, so the differences $15$ and $16$ are odd and even. A pair must contribute an odd exactly when its own sum is odd, which happens only once (the jump from $26$ to $41$). Hence exactly one odd is required — the same answer (A).
CCSS standards used (min grade 2)
2.NBT.B.5Fluently add and subtract within 100 (Recovering each pair's sum from the running totals ($41-26=15$, $57-41=16$) and adding the per-pair minimums $0+1+0=1$.)2.OA.C.3Determine whether a group of objects has an odd or even number (Extending odd/even to sums: a pair adds to an odd number only when exactly one member is odd, which forces the minimum count of odds in each pair.)
⭐ Two numbers only make an odd total when exactly one of them is odd, so this AMC 10 question is really a Grade 2 odd/even puzzle: only the pair that jumps by an odd amount is forced to carry a single odd number.
⭐ Two numbers only make an odd total when exactly one of them is odd, so this AMC 10 question is really a Grade 2 odd/even puzzle: only the pair that jumps by an odd amount is forced to carry a single odd number.
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