AMC 8 · 2017 · #12

Easy mode Grade 4

Problem

Imagine searching for a whole number bigger than $1$ that does the following:

When you divide it by $4$ , the remainder is $1$ .
When you divide it by $5$ , the remainder is $1$ .
When you divide it by $6$ , the remainder is $1$ .

Many numbers fit. We want the smallest one.

Find that smallest number. Which pair of numbers below does it sit between?

$\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124$

Pick an answer.

(A)

$2 ext{ and }19$

(B)

$20 ext{ and }39$

(C)

$40 ext{ and }59$

(D)

$60 ext{ and }79$

(E)

$80 ext{ and }124$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Find the smallest positive integer $N > 1$ such that dividing $N$ by $4$, by $5$, and by $6$ each leaves a remainder of $1$. Then say which of the given ranges contains $N$.

Givens: $N$ is a positive integer with $N > 1$; $N \div 4$ leaves remainder $1$; $N \div 5$ leaves remainder $1$; $N \div 6$ leaves remainder $1$; Answer choices: (A) $2$ and $19$, (B) $20$ and $39$, (C) $40$ and $59$, (D) $60$ and $79$, (E) $80$ and $124$

Unknowns: The smallest such $N$ (and the answer-choice range that contains it)

Understand

Restated: Find the smallest positive integer $N > 1$ such that dividing $N$ by $4$, by $5$, and by $6$ each leaves a remainder of $1$. Then say which of the given ranges contains $N$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

Three simultaneous remainder conditions feel intimidating, but Tool #9 transforms the problem into something much easier: "$N$ has remainder $1$ when divided by $d$" is the same as saying "$N - 1$ is a multiple of $d$." So instead of hunting for $N$ directly, we hunt for $N - 1$, which must be a common multiple of $4$, $5$, and $6$ — pure 4th-grade multiples thinking. Tool #2 (Systematic List) then sweeps multiples of the largest divisor ($6$) in order and checks each against the other two conditions, which is faster than computing a formal LCM. Tool #3 (Eliminate) closes the problem by matching $N$ against the five labeled ranges.

Execute — Answer: D

#9 Solve an Easier Related Problem 4.NBT.B.6 Step 1

Rewrite the three remainder conditions as one divisibility condition.
"$N$ leaves remainder $1$ when divided by $d$" means $N - 1$ is exactly divisible by $d$.
Applying this to $d = 4, 5, 6$ turns the problem into: find the smallest positive $N - 1$ that is a common multiple of $4$, $5$, and $6$.

$N - 1$ is a multiple of $4$, $5$, and $6$

💡 Remainder $1$ means "one past a multiple," so subtracting $1$ snaps the number back onto a multiple — a Grade 4 remainder idea.

#2 Make a Systematic List 4.OA.B.4 Step 2

List multiples of the largest divisor ($6$) in order and circle the ones that are also multiples of both $4$ and $5$.
Going through $6, 12, 18, 24, 30, 36, 42, 48, 54, 60$, only $60$ is divisible by $4$ (since $60 = 4 \times 15$) and by $5$ (since $60 = 5 \times 12$).
So the smallest positive common multiple of $4$, $5$, $6$ is $60$.

$6, 12, 18, 24, 30, 36, 42, 48, 54, \boxed{60}$ — $60 = 4 \times 15 = 5 \times 12 = 6 \times 10$

💡 Sweeping multiples of one number and checking divisibility by the others is the Grade 4 "factors and multiples" skill in action.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 3

Add $1$ to recover $N$.
Since the smallest valid value of $N - 1$ is $60$, the smallest valid $N$ is $60 + 1 = 61$.
(The next candidate would come from the next common multiple, $120$, giving $N = 121$ — much larger, so $61$ wins.)

$$N = 60 + 1 = 61$$

💡 Undoing the "$-1$" shift to translate between the easier problem and the original is exactly the Grade 4 multi-step word-problem move.

#3 Eliminate Possibilities 4.NBT.A.2 Step 4

Match $N = 61$ against the five range choices and eliminate the rest.
$61$ is not in $[2, 19]$, $[20, 39]$, or $[40, 59]$, and it is below $80$ so not in $[80, 124]$.
It fits cleanly into $[60, 79]$, which is choice (D).

$$60 < 61 < 79 \;\Rightarrow\; \textbf{(D)}$$

💡 Comparing $61$ to the range endpoints uses Grade 4 multi-digit number comparison.

[1] #9 4.NBT.B.6 Rewrite the three remainder conditions as one divisibility condition. "$N$ leave

[2] #2 4.OA.B.4 List multiples of the largest divisor ($6$) in order and circle the ones that ar

[3] #9 4.OA.A.3 Add $1$ to recover $N$. Since the smallest valid value of $N - 1$ is $60$, the s

[4] #3 4.NBT.A.2 Match $N = 61$ against the five range choices and eliminate the rest. $61$ is no

Review

Reasonableness: Quick check: $61 \div 4 = 15$ remainder $1$, $61 \div 5 = 12$ remainder $1$, $61 \div 6 = 10$ remainder $1$. All three remainder conditions hold. And $61$ is greater than $1$, so the "greater than 1" clause is satisfied. The only smaller candidate would come from a common multiple of $4$, $5$, $6$ that is smaller than $60$ — but our systematic list confirmed none exist. So $N = 61$ is genuinely the smallest, and it sits in the $60$–$79$ range, matching (D).

Alternative: Tool #6 (Guess and Check) directly on the answer choices: pick the smallest $N$ in each range and test the three remainders. (A) $N = 5$: $5 \div 4$ has remainder $1$, but $5 \div 5 = 1$ remainder $0$ — fail. (B) $N = 21$: $21 \div 4$ has remainder $1$, $21 \div 5$ has remainder $1$, but $21 \div 6 = 3$ remainder $3$ — fail. (C) $N = 41$: $41 \div 4$ has remainder $1$, $41 \div 5$ has remainder $1$, but $41 \div 6 = 6$ remainder $5$ — fail. (D) $N = 61$: all three give remainder $1$ — works. So (D).

CCSS standards used (min grade 4)

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Reading the phrase "leaves a remainder of $1$" as the divisibility statement "$N - 1$ is exactly divisible by $d$" — the Grade 4 remainder concept made explicit.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing multiples of $6$ in order and checking each one for divisibility by $4$ and $5$ until the first common multiple ($60$) appears.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Translating between the helper quantity $N - 1$ and the original $N$ by adding $1$ at the end — the multi-step bookkeeping of the word problem.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Comparing $N = 61$ against the endpoints of the five labeled ranges ($19, 39, 59, 79, 124$) to choose the correct interval.)

⭐ This AMC 8 problem only needs Grade 4 multiples and remainders you already know — "remainder 1" just means "one more than a multiple"!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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