AMC 10 · 2002 · #4
Grade 6 arithmeticPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The phrase 'there exists at least one n' hands you a choice: for each m you get to pick the n that helps most. Tool #6 (Guess and Check) turns that freedom into a plan — instead of testing many n, guess the single simplest one, n=1, and check what the inequality becomes. Tool #4 (Introduce a Variable) keeps m general so one check covers every value at once rather than one number at a time. Tool #3 (Eliminate Possibilities) then reads the multiple-choice list: the four finite counts (A)-(D) can only survive if some m fails, so showing that none fail rules them all out and leaves 'infinitely many'.
Pick the easiest n
Since n is yours to choose, take the smallest one, n=1 — the two sides become m · 1=m and m+1.
Since only one n has to work, use the freedom to pick the friendliest value instead of hunting.
6.EE.A.2Guess And CheckCheck it holds for every m
m ≤ m+1 just says a number is no bigger than itself plus one, so n=1 works for every m.
Adding one to a number can never make it smaller, so m ≤ m+1 can never fail.
Adding one to a number can never make it smaller, so the comparison can never fail.
▸ Why?
A quantity is always below the same quantity increased, so the order is fixed for every value.
▸ Why?
Since no single value can break it, no counterexample exists and the claim survives everywhere.
Count the winners
Every m pairs with n=1 and the positive integers never run out, so no finite count survives — answer (E).
If no m can be excluded, the count cannot be any finite number.
6.EE.B.5Eliminate PossibilitiesWhen a problem only needs one value to exist, you get to choose it — here picking n=1 makes the inequality m ≤ m+1, which is always true, so every m works.
- Pick the easiest n
- Check it holds for every m
- Count the winners