Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #19
Grade 6 number-theoryPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Squaring then subtracting sounds heavy, so start with Tool #9 (Easier Related Problem): try the smallest consecutive pairs and just compute. The numbers fall into a clean pattern, which is Tool #5 (Look for a Pattern) — the difference of squares of consecutive integers equals the sum of those integers, so it is always odd and at most 99. Tool #3 (Eliminate Possibilities) then crosses out every choice that fails the odd-and-less-than-100 filter. We do not need Tool #13 (Algebra) to set up an equation; the pattern from a few small cases is doing all the work.
Try small consecutive pairs
Shrink the problem (Tool #9): compute the difference of squares for small pairs — 3, 5, 7, 9.
Squaring sounds intimidating, but tiny pairs make the arithmetic trivial and show the structure right away.
6.EE.A.1Solve An Easier Related ProblemSpot the pattern
Spot the pattern: those odd results are exactly the pair's own sum, so the difference of squares equals 2n + 1.
Whatever you put in, the answer always comes out as twice n plus one — an odd number, and the same as n + (n+1).
For two consecutive integers n and n+1, the difference of their squares (n+1)² - n² equals the two integers added together, 2n + 1.
▸ Why?
Squaring the larger integer gives (n+1)² = n² + 2n + 1, which is the smaller square with an extra 2n + 1 stacked on top.
▸ Why?
Multiplying (n+1) by (n+1) spreads each part of one factor across each part of the other, n · n + n · 1 + 1 · n + 1 · 1, which collects into n² + 2n + 1.
▸ Why?
Subtracting the smaller square n² from n² + 2n + 1 takes back exactly the n² that was there, leaving only 2n + 1.
▸ Why?
The remaining 2n + 1 is the same thing as n + (n+1), the two consecutive integers added together, because the two copies of n merge into 2n.
▸ Why?
2n just means two equal groups of n, which is n added to n, so n + (n+1) regroups into 2n + 1.
Turn the pattern into filters
Read off two filters: the difference 2n + 1 is always odd, and it stays under 100.
2n is even, so adding 1 always lands on odd. And 2n+1 ＜ 100 comes straight from the problem's "sum less than 100" condition.
6.NS.B.4Look For A PatternFilter the answer choices
Apply both filters (Tool #3): 2, 64, 96 are even; 131 tops 99 — only 79 survives, odd and below 100.
Two quick tests (odd? under 100?) crush four of the five choices in one pass.
6.EE.B.5Eliminate PossibilitiesFind the pair that works
Confirm it's reachable: 2n + 1 = 79 gives n = 39, and 40² - 39² = 79. The answer is (C).
Showing the actual pair (39, 40) proves 79 is achievable, not just "not yet eliminated".
6.EE.B.7Eliminate PossibilitiesThe difference of squares of two consecutive integers is just their sum — always odd, and here below 100. That one fact knocks out four choices and lands on (C) 79.
- Try small consecutive pairs
- Spot the pattern
- Turn the pattern into filters
- Filter the answer choices
- Find the pair that works
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