AMC 8 · 2012 · #15

Easy mode Grade 6

Problem

Find a whole number that is greater than $2$ and has this special property: when you divide it by $3$ , the remainder is $2$ . When you divide it by $4$ , the remainder is $2$ . When you divide it by $5$ , the remainder is $2$ . When you divide it by $6$ , the remainder is also $2$ .

There are many such numbers. We want the smallest one.

Which of these ranges does that smallest number fall into?

Pick an answer.

(A)

$hspace{.05in}40 ext{ and }50$

(B)

$hspace{.05in}51 ext{ and }55$

(C)

$hspace{.05in}56 ext{ and }60$

(D)

$hspace{.05in}61 ext{ and }65$

(E)

$hspace{.05in}66 ext{ and }99$

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Toolkit + CCSS Solution

Understand

Restated: Find the smallest whole number bigger than $2$ that leaves a remainder of $2$ when you divide it by $3$, by $4$, by $5$, and by $6$. Then say which of the given ranges it falls inside.

Givens: The number $N$ we want is greater than $2$; $N$ divided by $3$ leaves remainder $2$; $N$ divided by $4$ leaves remainder $2$; $N$ divided by $5$ leaves remainder $2$; $N$ divided by $6$ leaves remainder $2$; Answer choices (ranges): (A) $40$–$50$, (B) $51$–$55$, (C) $56$–$60$, (D) $61$–$65$, (E) $66$–$99$

Unknowns: The smallest such $N$; Which answer-choice range contains that $N$

Understand

Restated: Find the smallest whole number bigger than $2$ that leaves a remainder of $2$ when you divide it by $3$, by $4$, by $5$, and by $6$. Then say which of the given ranges it falls inside.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities

The problem packs four divisibility conditions into one question. Tool #7 (Identify Subproblems) turns that into one clean step: subtract the common remainder of $2$, and the question becomes "what is the smallest positive number that is divisible by $3$, $4$, $5$, and $6$ at the same time?" — the definition of the least common multiple. Tool #5 (Look for a Pattern) handles the leftover comparison: every working $N$ has the form $60k + 2$, so we just pick the smallest $k$ and walk the pattern. Tool #3 (Eliminate Possibilities) then maps the resulting $N$ against the five range choices and rules out the four that miss. We avoid tool #13 (Algebra) and modular-arithmetic notation because the "shift by $2$" subproblem move makes them unnecessary.

Execute — Answer: D

#7 Identify Subproblems 4.OA.B.4 Step 1

Pull the constant remainder out.
Saying "$N$ leaves remainder $2$ when divided by $d$" is the same as saying "$N - 2$ is a multiple of $d$".
Doing this for all four divisors at once turns the original problem into a much simpler one: find the smallest positive value of $N - 2$ that is a multiple of $3$, $4$, $5$, and $6$.

$$N \equiv 2 \;(\bmod\; 3,4,5,6) \;\Longleftrightarrow\; N - 2 \text{ is a multiple of each of } 3,4,5,6$$

💡 Stripping the shared remainder is the Tool #7 subproblem move — one harder question becomes one easier question about plain multiples.

#7 Identify Subproblems 6.NS.B.4 Step 2

Find the smallest positive number that is a multiple of $3$, $4$, $5$, and $6$ — the least common multiple.
Use prime factorization: $3 = 3$, $4 = 2^2$, $5 = 5$, $6 = 2 \times 3$.
Take the highest power of each prime that shows up.

$$\mathrm{LCM}(3,4,5,6) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$$

💡 The $6$ is "free" because $6 = 2 \times 3$ is already covered by the $4$ and $3$ — no new prime is added.

#5 Look for a Pattern 4.OA.C.5 Step 3

List the candidate values of $N$ in order and pick the first one that is greater than $2$.
Since $N - 2$ must be a positive multiple of $60$, write $N - 2 = 60k$ for $k = 1, 2, 3, \dots$ and read off the pattern.

$$k = 1 \Rightarrow N = 62; \;\; k = 2 \Rightarrow N = 122; \;\; k = 3 \Rightarrow N = 182; \;\dots$$

💡 Listing the first few terms of the "add $60$" pattern makes the smallest one — and the gap to the next one — obvious.

#3 Eliminate Possibilities 1.NBT.B.3 Step 4

The smallest valid $N$ is $62$.
Check the ranges in the answer choices: $40$–$50$, $51$–$55$, $56$–$60$, $61$–$65$, $66$–$99$.
Only one of them contains $62$.

$$61 \le 62 \le 65 \;\Rightarrow\; \textbf{(D)}$$

💡 Comparing $62$ against each range is just the Tool #3 "eliminate possibilities" move — four ranges miss, one fits.

[1] #7 4.OA.B.4 Pull the constant remainder out. Saying "$N$ leaves remainder $2$ when divided b

[2] #7 6.NS.B.4 Find the smallest positive number that is a multiple of $3$, $4$, $5$, and $6$ —

[3] #5 4.OA.C.5 List the candidate values of $N$ in order and pick the first one that is greater

[4] #3 1.NBT.B.3 The smallest valid $N$ is $62$. Check the ranges in the answer choices: $40$–$50

Review

Reasonableness: Verify $N = 62$ directly: $62 = 3 \times 20 + 2$ (remainder $2$, good), $62 = 4 \times 15 + 2$ (good), $62 = 5 \times 12 + 2$ (good), $62 = 6 \times 10 + 2$ (good). All four conditions hold. And nothing smaller works, because any smaller candidate $N - 2$ would have to be a positive multiple of $60$ smaller than $60$ — impossible. So $62$ is the smallest, and $62$ sits in the $61$–$65$ range, confirming (D).

Alternative: Tool #3 (Eliminate Possibilities) directly on the choices: any valid $N$ has the form $60k + 2$, so the only options inside the listed ranges are $62$ (range D) and $122$ (range E, but $122 > 99$, so E is also out for the smallest one). Range A ($40$–$50$), B ($51$–$55$), and C ($56$–$60$) cannot contain any number of the form $60k + 2$ at all, since $60(0) + 2 = 2$ is too small and $60(1) + 2 = 62$ is already past them. Only (D) survives.

CCSS standards used (min grade 6)

1.NBT.B.3 Compare two two-digit numbers using <, =, > symbols (Checking that $62$ lies between $61$ and $65$ to pick the correct answer-choice range.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that "leaves remainder $2$ when divided by $d$" is the same as "$N - 2$ is a multiple of $d$", and stating this for all four divisors at once.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Generating the list of valid $N$ values $62, 122, 182, \dots$ by repeatedly adding $60$ to $2$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Computing $\mathrm{LCM}(3, 4, 5, 6) = 60$ via prime factorization — the key number that drives the whole solution.)

⭐ This AMC 8 problem boils down to one Grade 6 idea — finding the least common multiple of $3, 4, 5, 6$ — once you spot that subtracting the shared remainder of $2$ makes the four conditions collapse into one!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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