AMC 10 · 2003 · #3
Grade 7 algebraThe sum of 5 consecutive even integers is 4 less than the sum of the first 8 consecutive odd counting numbers. What is the smallest of the even integers?
Pick an answer.
AMC 10 2003 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: Five consecutive even integers add up to a total that is $4$ less than the sum of the first $8$ odd counting numbers. Find the smallest of the five even integers.
Givens: There are $5$ consecutive even integers (each $2$ more than the one before).; Their sum is $4$ less than the sum of the first $8$ consecutive odd counting numbers ($1,3,5,7,\dots$).; Answer choices: (A) $6$, (B) $8$, (C) $10$, (D) $12$, (E) $14$
Unknowns: The smallest of the $5$ consecutive even integers.
Understand
Restated: Five consecutive even integers add up to a total that is $4$ less than the sum of the first $8$ odd counting numbers. Find the smallest of the five even integers.
Givens: There are $5$ consecutive even integers (each $2$ more than the one before).; Their sum is $4$ less than the sum of the first $8$ consecutive odd counting numbers ($1,3,5,7,\dots$).; Answer choices: (A) $6$, (B) $8$, (C) $10$, (D) $12$, (E) $14$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #5 Look for a Pattern, #13 Convert to Algebra
The five unknown evens are all tied to one number — the smallest — so naming it $x$ (Tool #4) turns five mystery values into one expression $5x+20$. Tool #5 (Look for a Pattern) collapses the other side fast: the first $n$ odd numbers always total $n^2$, so eight of them give $64$ without adding one by one. Tool #13 (Convert to Algebra) then turns the sentence '$4$ less than' into a single equation to solve.
Execute — Answer: B
4.OA.C.5 Step 1 Add the first 8 odd numbers
- The first $8$ odd counting numbers are $1,3,5,7,9,11,13,15$.
- You can add them straight across to get $64$.
- Faster: the first $n$ odd numbers always total $n^2$, so the first $8$ total $8^2=64$.
💡 Odd numbers stacked from $1$ build a perfect square, so $n$ of them make an $n\times n$ block.
6.EE.B.6 Step 2 Name the smallest even integer
- Let $x$ stand for the smallest of the five even integers.
- Because consecutive even integers jump by $2$, the five are $x,\ x+2,\ x+4,\ x+6,\ x+8$.
- Adding them, the $x$'s give $5x$ and the numbers give $2+4+6+8=20$, so the total is $5x+20$.
💡 One name for the smallest value forces every other value, so five unknowns shrink to one.
7.EE.B.4 Step 3 Turn the sentence into an equation
- The even sum is '$4$ less than' the odd sum of $64$, which means $64-4=60$.
- Setting the even total equal to that gives the equation $5x+20=60$.
💡 '$4$ less than $64$' is just $64-4$, and the word 'is' becomes an equals sign.
7.EE.B.4 Step 4 Solve and check
- Subtract $20$ from both sides: $5x=40$.
- Divide by $5$: $x=8$.
- So the smallest even integer is $8$, which is choice (B).
- Check: the five evens $8,10,12,14,16$ add to $60$, and $60$ is indeed $4$ less than $64$.
💡 Undo the equation in reverse — remove the $+20$, then the $\times5$ — to uncover $x$.
4.OA.C.5 The first $8$ odd counting numbers are $1,3,5,7,9,11,13,15$. You can add them st 6.EE.B.6 Let $x$ stand for the smallest of the five even integers. Because consecutive ev 7.EE.B.4 The even sum is '$4$ less than' the odd sum of $64$, which means $64-4=60$. Sett 7.EE.B.4 Subtract $20$ from both sides: $5x=40$. Divide by $5$: $x=8$. So the smallest ev Review
Reasonableness: The answer $x=8$ sits near the middle of the choices, which is a good sign. Plugging back, $8+10+12+14+16=60$, and $64-60=4$, exactly the '$4$ less' the problem asked for. If you had mistakenly subtracted $4$ from the even side, or forgotten the $+20$ from the constants, you would land on one of the other choices — so the back-check is what confirms (B).
Alternative: Skip the variable and use the average. Five consecutive even integers are symmetric about their middle value, so their sum is $5$ times the middle number. The sum is $60$, so the middle even integer is $60\div5=12$. The smallest is two steps of $2$ below the middle: $12-4=8$.
CCSS standards used (min grade 7)
4.OA.C.5Generate and analyze a number or shape pattern that follows a given rule (Recognizing that the first $8$ odd counting numbers sum to $8^2=64$.)6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Letting $x$ be the smallest even integer and writing the five as $x,x+2,x+4,x+6,x+8$ with sum $5x+20$.)7.EE.B.4Construct and solve simple equations of the form $px+q=r$ to solve problems (Setting up $5x+20=60$ and solving it to get $x=8$.)
⭐ Name the smallest number $x$, write the rest as steps of $2$ above it, and one clean equation does the rest.
⭐ Name the smallest number $x$, write the rest as steps of $2$ above it, and one clean equation does the rest.
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