AMC 10 · 2011 · #4
Grade 4 countingLet X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of Y−X?
Pick an answer.
AMC 10 2011 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: Two lists of even numbers are added up: $X = 10+12+14+\cdots+100$ and $Y = 12+14+16+\cdots+102$. Find how much bigger the second total is than the first, that is $Y-X$.
Givens: $X$ adds the even numbers from $10$ up to $100$: $X = 10+12+14+\cdots+100$; $Y$ adds the even numbers from $12$ up to $102$: $Y = 12+14+16+\cdots+102$; Answer choices: (A) $92$, (B) $98$, (C) $100$, (D) $102$, (E) $112$
Unknowns: The value of the difference $Y-X$
Understand
Restated: Two lists of even numbers are added up: $X = 10+12+14+\cdots+100$ and $Y = 12+14+16+\cdots+102$. Find how much bigger the second total is than the first, that is $Y-X$.
Givens: $X$ adds the even numbers from $10$ up to $100$: $X = 10+12+14+\cdots+100$; $Y$ adds the even numbers from $12$ up to $102$: $Y = 12+14+16+\cdots+102$; Answer choices: (A) $92$, (B) $98$, (C) $100$, (D) $102$, (E) $112$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
Adding up $46$ numbers twice would be slow and error-prone. Tool #5 (Look for a Pattern) spots the real structure instead: each number in $Y$ is exactly $2$ more than the matching number in $X$, so the whole difference is just (number of terms) times $2$. Tool #7 (Identify Subproblems) handles the one supporting job — counting how many terms there are. Tool #16 (Change Focus) gives a fast cross-check by lining the two sums up and cancelling the block they share.
Execute — Answer: A
4.OA.C.5 Step 1 Match the lists term by term
- Write the two sums so their terms line up: $X = 10,12,14,\dots,100$ and $Y = 12,14,16,\dots,102$.
- Pair the first term of $X$ with the first term of $Y$, the second with the second, and so on.
- Every term of $Y$ is exactly $2$ more than the term it is paired with in $X$: $12-10=2$, $14-12=2$, $16-14=2$, and so on.
💡 Both lists climb by the same step, so shifting the start by one step makes every paired number differ by exactly that step.
4.OA.A.3 Step 2 Count how many terms there are
- $X$ lists the even numbers from $10$ to $100$.
- Count them: the span is $100-10 = 90$, divide by the step to get $90 \div 2 = 45$, then add $1$ for the starting number, giving $45+1 = 46$ terms.
- $Y$ also has $46$ terms, so there are $46$ matched pairs.
💡 To count evenly-spaced numbers, see how many steps fit between the ends, then add one for the number you started on.
4.NBT.B.5 Step 3 Add up the differences
- Each of the $46$ pairs differs by $2$, and $Y-X$ is just the total of all those little differences.
- So $Y-X = 46 \times 2 = 92$.
- That matches choice (A).
💡 If every pair is ahead by the same amount, the whole list is ahead by that amount times the number of pairs.
4.OA.C.5 Write the two sums so their terms line up: $X = 10,12,14,\dots,100$ and $Y = 12, 4.OA.A.3 $X$ lists the even numbers from $10$ to $100$. Count them: the span is $100-10 = 4.NBT.B.5 Each of the $46$ pairs differs by $2$, and $Y-X$ is just the total of all those Review
Reasonableness: The difference should be small compared with the totals, because the two lists are almost identical — only their ends differ. A per-term gap of $2$ across $46$ terms lands at $92$, which is choice (A). The larger choices ($98, 100, 102, 112$) would each need a per-term gap bigger than $2$ or more than $46$ terms, and neither is true.
Alternative: Line the sums up and cancel (change focus, Tool #16): the block $12+14+\cdots+100$ appears in both $X$ and $Y$, so it cancels in $Y-X$. What is left is the $102$ that only $Y$ has minus the $10$ that only $X$ has: $Y-X = 102-10 = 92$, confirming (A).
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Spotting that each term of $Y$ is $2$ more than its matching term in $X$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Counting the terms with $(100-10) \div 2 + 1 = 46$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Multiplying $46 \times 2 = 92$ to total the per-pair differences.)
⭐ When two lists are the same except shifted over, don't add them up — just multiply how far each number moved by how many numbers there are.
⭐ When two lists are the same except shifted over, don't add them up — just multiply how far each number moved by how many numbers there are.
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