AMC 10 · 2012 · #5
Easy mode Grade 2Integers are the numbers …,−2,−1,0,1,2,… — no fractions, and negatives are allowed. Mina writes down six integers, two at a time. The first two add up to 26. Once the next two are written, all four together add up to 41. Once the last two are written, all six together add up to 57. What is the smallest number of even integers Mina's six numbers could include?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Six integers are handed over two at a time. After the first two the total is $26$, after four integers the total is $41$, and after all six the total is $57$. Find the smallest number of even integers the six can possibly contain.
Givens: The first two integers have sum $26$; Two more integers are added and the running total becomes $41$; Two more integers are added again and the running total becomes $57$; All six numbers are integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The smallest possible count of even numbers among the six integers
Understand
Restated: Six integers are handed over two at a time. After the first two the total is $26$, after four integers the total is $41$, and after all six the total is $57$. Find the smallest number of even integers the six can possibly contain.
Givens: The first two integers have sum $26$; Two more integers are added and the running total becomes $41$; Two more integers are added again and the running total becomes $57$; All six numbers are integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement, #6 Guess and Check
The question asks for a minimum, so Tool #14 (Extreme Principle) sets the shape of the work: find a floor no set can go below, then produce one set that sits exactly on the floor. Tool #7 (Identify Subproblems) makes the floor findable — the running totals cut the six integers into three independent pairs, each with its own sum. Tool #16 (Change Focus) drops the actual sizes and keeps only even-or-odd, which is all the count depends on. Tool #6 (Guess and Check) then builds a concrete list of six integers that hits every stated total while carrying as few even numbers as the floor allows.
Execute — Answer: A
2.OA.A.1 Step 1 Cut the six into three pairs
- The integers arrive two at a time, so the running totals separate them.
- The first pair sums to $26$.
- The first four sum to $41$, so the second pair added $41-26=15$.
- All six sum to $57$, so the third pair added $57-41=16$.
- The problem is now three separate pairs with required sums $26$, $15$, and $16$, and each pair may be chosen freely as long as it hits its own sum.
💡 Two running totals differ by exactly what was added in between, so subtracting them isolates each new pair.
2.OA.C.3 Step 2 Watch parity, not size
- How big the integers are does not matter — only whether each one is even or odd.
- Two odd numbers add to an even number, two even numbers add to an even number, and one of each adds to an odd number.
- Read those backwards.
- A pair whose sum is odd has to be one even and one odd, so it always contains exactly one even number.
- A pair whose sum is even is either two odds or two evens, so it is allowed to contain no even number at all.
- Each pair's own sum therefore decides how cheap that pair can be.
💡 Even and odd combine in a fixed pattern, so a pair's sum reports back how many of its two numbers are even.
2.OA.C.3 Step 3 Prove zero evens is impossible
- Apply that rule to each pair.
- The sums $26$ and $16$ are even, so those two pairs are permitted to be two odd numbers each and can contribute no even integers.
- The middle pair sums to $15$, which is odd.
- It cannot be two odd numbers, because two odds always total an even number, and it cannot be two even numbers for the same reason.
- It must be one even and one odd.
- So at least one even integer appears among the six no matter how the numbers are chosen — a count of $0$ is impossible, and $1$ is the lowest count still open.
💡 A single odd pair-sum forces an even number into the list, and it forces only one.
2.NBT.B.5 Step 4 Build a list that reaches one
- A floor only becomes the answer if some real set of six integers lands on it.
- Take $15$ and $11$ for the first pair, both odd and summing to $26$.
- Take $8$ and $7$ for the second, summing to $15$ with a single even number.
- Take $9$ and $7$ for the third, both odd and summing to $16$.
- Follow the story to check: $15+11=26$, then $26+8+7=41$, then $41+9+7=57$.
- Every stated total is hit, and exactly one of the six numbers, the $8$, is even.
- One even integer is both unavoidable and enough, so the minimum is $1$; the larger counts in (B) through (E) are all beaten by this list.
- The answer is (A).
💡 A bound you can actually land on is the true minimum; a bound nothing reaches is only a guess.
2.OA.A.1 The integers arrive two at a time, so the running totals separate them. The firs 2.OA.C.3 How big the integers are does not matter — only whether each one is even or odd. 2.OA.C.3 Apply that rule to each pair. The sums $26$ and $16$ are even, so those two pair 2.NBT.B.5 A floor only becomes the answer if some real set of six integers lands on it. Ta Review
Reasonableness: Check the witness against the original wording rather than against the pairs. The six numbers $15$, $11$, $8$, $7$, $9$, $7$ give $15+11=26$ for the first two, then $26+8+7=41$ once two more are added, then $41+9+7=57$ once the last two are added, so all three stated totals are exact. Counting even numbers in that list gives one, the $8$. Step 3 already showed zero is impossible, so the minimum is pinned between $1$ and $1$. Both halves of the minimum claim hold, so (A) is consistent.
Alternative: Read parity straight off the running totals and never form the pair sums. Suppose the six contained no even integer, so all six are odd. Two odd numbers total an even number, four odd numbers total an even number, and six odd numbers total an even number, so all three running totals would have to be even. The second running total is $41$, which is odd, so the assumption collapses and at least one of the first four integers is even. The list $15$, $11$, $8$, $7$, $9$, $7$ then shows one even integer is enough, giving the same minimum of $1$ without ever computing $15$ or $16$.
CCSS standards used (min grade 2)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Turning the three running totals into the three pair sums $26$, $15$, and $16$.)2.OA.C.3Determine whether a group of objects has an odd or even number (Using the even-odd rules for sums to show a pair summing to $15$ must hold exactly one even number, while pairs summing to $26$ and $16$ may hold none.)2.NBT.B.5Fluently add and subtract within 100 (Checking that $15$, $11$, $8$, $7$, $9$, $7$ produces the running totals $26$, $41$, and $57$.)
⭐ Two odd numbers always add up to an even number, so any pair that must add to an odd total is forced to hide exactly one even number inside it.
⭐ Two odd numbers always add up to an even number, so any pair that must add to an odd total is forced to hide exactly one even number inside it.
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