AMC 10 · 2002 · #4
Grade 6 arithmeticFor how many positive integers m does there exist at least one positive integer n such that m⋅n≤m+n?
Pick an answer.
AMC 10 2002 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: Count the positive integers $m$ for which you can find at least one positive integer $n$ making $m\cdot n\le m+n$ true.
Givens: The inequality $m\cdot n\le m+n$; Both $m$ and $n$ must be positive integers; You only need one working $n$ for a given $m$ to count that $m$; Answer choices: (A) $4$, (B) $6$, (C) $9$, (D) $12$, (E) infinitely many
Unknowns: How many positive integers $m$ have at least one valid partner $n$
Understand
Restated: Count the positive integers $m$ for which you can find at least one positive integer $n$ making $m\cdot n\le m+n$ true.
Givens: The inequality $m\cdot n\le m+n$; Both $m$ and $n$ must be positive integers; You only need one working $n$ for a given $m$ to count that $m$; Answer choices: (A) $4$, (B) $6$, (C) $9$, (D) $12$, (E) infinitely many
Plan
Primary tool: #6 Guess and Check
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
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'.
Execute — Answer: E
6.EE.A.2 Step 1 Pick the easiest n
- You are promised a free choice of $n$, so choose the smallest positive integer, $n=1$.
- Substituting $n=1$ turns the two sides into $m\cdot 1=m$ and $m+1$.
- The whole question shrinks to one clean comparison: is $m\le m+1$?
💡 Since only one $n$ has to work, use the freedom to pick the friendliest value instead of hunting.
6.EE.B.5 Step 2 Check it holds for every m
- The inequality $m\le m+1$ says a number is no bigger than itself plus one.
- That is true for every positive integer $m$ — adding $1$ always makes a number larger.
- So $n=1$ works no matter which $m$ you started with; not a single $m$ is left out.
💡 Adding one to a number can never make it smaller, so $m\le m+1$ can never fail.
6.EE.B.5 Step 3 Count the winners
- Every positive integer $m$ has a working partner (namely $n=1$), and there are infinitely many positive integers, so infinitely many values of $m$ qualify.
- That kills the four finite counts: (A) $4$, (B) $6$, (C) $9$, and (D) $12$ would each require some $m$ to fail, but none do.
- The answer is (E).
💡 If no $m$ can be excluded, the count cannot be any finite number.
6.EE.A.2 You are promised a free choice of $n$, so choose the smallest positive integer, 6.EE.B.5 The inequality $m\le m+1$ says a number is no bigger than itself plus one. That 6.EE.B.5 Every positive integer $m$ has a working partner (namely $n=1$), and there are i Review
Reasonableness: Spot-check a few values with $n=1$: $m=5$ gives $5\le 6$, $m=100$ gives $100\le 101$, $m=1$ gives $1\le 2$ — all true, and there is no largest $m$ to try, so the count really is unbounded. It also makes sense that a finite answer like $4$ or $12$ would be suspicious here: the inequality gives you a free, always-winning move ($n=1$), so there is no reason it should ever stop working, which is exactly what 'infinitely many' captures.
Alternative: Rewrite algebraically: $m n\le m+n$ becomes $mn-m-n+1\le 1$, i.e. $(m-1)(n-1)\le 1$. Choosing $n=1$ makes the left side $(m-1)\cdot 0=0\le 1$ for every $m$, confirming that all positive integers $m$ work and the answer is (E). This factored form also shows why $n=1$ is the magic choice — it zeroes out the product no matter how big $m$ is.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $n=1$ into $m\cdot n$ and $m+n$ to reduce the inequality to $m\le m+1$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Deciding which positive integers $m$ satisfy $m\le m+1$ (all of them) and concluding the count is infinite.)
⭐ When a problem only needs one value to exist, you get to choose it — here picking $n=1$ makes the inequality $m\le m+1$, which is always true, so every $m$ works.
⭐ When a problem only needs one value to exist, you get to choose it — here picking $n=1$ makes the inequality $m\le m+1$, which is always true, so every $m$ works.
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