AMC 10 · 2012 · #22
Grade 8 arithmeticThe sum of the first m positive odd integers is 212 more than the sum of the first n positive even integers. What is the sum of all possible values of n?
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: Add up the first m odd numbers, and separately add up the first n even numbers. The odd total is 212 bigger than the even total. Find every whole number n that makes this possible, then add those n-values together.
Givens: The sum of the first m positive odd integers (1, 3, 5, ...) is being compared.; The sum of the first n positive even integers (2, 4, 6, ...) is the other quantity.; The odd sum is exactly 212 more than the even sum.; m and n are positive whole numbers.
Unknowns: All possible values of n.; The sum of those possible n-values.
Understand
Restated: Add up the first m odd numbers, and separately add up the first n even numbers. The odd total is 212 bigger than the even total. Find every whole number n that makes this possible, then add those n-values together.
Givens: The sum of the first m positive odd integers (1, 3, 5, ...) is being compared.; The sum of the first n positive even integers (2, 4, 6, ...) is the other quantity.; The odd sum is exactly 212 more than the even sum.; m and n are positive whole numbers.
Plan
Primary tool: #13 Convert to Algebra
Secondary: #4 Introduce a Variable, #5 Look for a Pattern, #2 Make a Systematic List, #3 Eliminate Possibilities
The two sums have clean closed forms, so the whole word problem becomes one equation. The trick is to reshape that equation into a difference of two squares; then the leftover number 847 only has a few factor pairs, and each pair points to one value of n. So the path is: turn the sums into formulas, build the equation, factor it, and list the finite possibilities.
Execute — Answer: A
4.OA.C.5 Step 1 Turn each sum into a formula
- The first odd numbers pile up to perfect squares: 1 = 1, 1+3 = 4, 1+3+5 = 9, so the sum of the first m odd numbers is m squared.
- The first even numbers pile up differently: 2 = 1x2, 2+4 = 2x3, 2+4+6 = 3x4, so the sum of the first n even numbers is n(n+1).
💡 Adding a whole run of odds or evens follows a fixed rule, so each sum collapses into one short formula.
6.EE.B.6 Step 2 Write the equation
- The odd sum is 212 more than the even sum, so set the odd formula equal to the even formula plus 212: m squared = n(n+1) + 212.
- Multiply out the right side to get m squared = n squared + n + 212.
💡 Naming the two counts m and n turns the whole sentence into one exact equation.
8.EE.A.2 Step 3 Reshape into a difference of squares
- The single n is awkward, so multiply the whole equation by 4 to force a perfect square to appear: 4m squared = 4n squared + 4n + 848.
- Since 4n squared + 4n + 1 = (2n+1) squared, the terms 4n squared + 4n equal (2n+1) squared minus 1.
- That leaves 4m squared = (2n+1) squared + 847.
- Moving the square across gives (2m) squared minus (2n+1) squared = 847, a difference of two squares.
💡 Multiplying by 4 squeezes a perfect square out of the messy n-terms, exposing a clean difference of squares.
4.OA.B.4 Step 4 Factor and list the divisor pairs
- A difference of two squares splits into a product: (2m - (2n+1))(2m + (2n+1)) = 847.
- Because m and n are positive, both factors are positive whole numbers and the second is the larger one, so they form a factor pair of 847.
- Since 847 = 7 x 11 x 11, its factor pairs are 1 x 847, 7 x 121, and 11 x 77.
💡 Once two whole numbers multiply to 847, the answer must hide inside one of 847's few factor pairs.
8.EE.C.7 Step 5 Solve each pair for n and add
- For any pair, the larger factor minus the smaller equals (2m+2n+1) - (2m-2n-1) = 4n + 2.
- From 1 and 847: 4n+2 = 846, so n = 211.
- From 7 and 121: 4n+2 = 114, so n = 28.
- From 11 and 77: 4n+2 = 66, so n = 16.
- Each pair also gives a whole-number m (212, 32, 22), so all three values of n are allowed.
- Adding them: 211 + 28 + 16 = 255, which is answer (A).
💡 Subtracting the paired factors cancels m and isolates n, so each factor pair hands over exactly one value of n.
4.OA.C.5 The first odd numbers pile up to perfect squares: 1 = 1, 1+3 = 4, 1+3+5 = 9, so 6.EE.B.6 The odd sum is 212 more than the even sum, so set the odd formula equal to the e 8.EE.A.2 The single n is awkward, so multiply the whole equation by 4 to force a perfect 4.OA.B.4 A difference of two squares splits into a product: (2m - (2n+1))(2m + (2n+1)) = 8.EE.C.7 For any pair, the larger factor minus the smaller equals (2m+2n+1) - (2m-2n-1) = Review
Reasonableness: Check the biggest case: n = 211 gives an even sum of 211 x 212 = 44732, and m = 212 gives an odd sum of 212 squared = 44944; the difference is 44944 - 44732 = 212, exactly as required. The cases n = 28 (m = 32) and n = 16 (m = 22) check the same way. Three valid n-values add to 255, matching choice (A); the nearby choices 256 through 259 would each need an extra or shifted solution that the factor pairs of 847 simply do not provide.
Alternative: Skip the factoring and test values directly: for each n, compute n^2 + n + 212 and check whether it is a perfect square m^2. It works only at n = 16 (484 = 22^2), n = 28 (1024 = 32^2), and n = 211 (44944 = 212^2); past that the gap between consecutive squares grows too fast to catch another, so the sum is again 16 + 28 + 211 = 255.
CCSS standards used (min grade 8)
4.OA.C.5Generate a number or shape pattern that follows a given rule (Recognizing that the first m odd numbers sum to m^2 and the first n even numbers sum to n(n+1).)6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the counts m and n and writing the equation m^2 = n^2 + n + 212 from the word problem.)8.EE.A.2Use square root symbols and work with perfect squares (Multiplying by 4 and completing the square to rewrite the equation as (2m)^2 - (2n+1)^2 = 847.)4.OA.B.4Find all factor pairs for a whole number and recognize multiples of each factor (Listing every factor pair of 847 (1x847, 7x121, 11x77) that the two positive factors could equal.)8.EE.C.7Solve linear equations in one variable (Using 4n + 2 equal to the difference of each factor pair to solve for n = 211, 28, 16 and summing them.)
⭐ Turn 'odds add to a square' and 'evens add to n(n+1)' into one equation, multiply by 4 to make a difference of squares, and the few factor pairs of 847 reveal every possible n.
⭐ Turn 'odds add to a square' and 'evens add to n(n+1)' into one equation, multiply by 4 to make a difference of squares, and the few factor pairs of 847 reveal every possible n.
More like this
Same archetype — closest grade level first.