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AMC 8 · 2017 · #7

Easy mode Grade 4
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Problem

Picture a 66-digit number where the first three digits and the last three digits are exactly the same, in the same order.

For example, 247247247247 works: the first three digits are 247247, and so are the last three. Another one is 358358358358.

Call any such 66-digit number ZZ.

Among the five choices below, one number is guaranteed to be a factor of every such ZZ. Which one?

(A) 11(B) 19(C) 101(D) 111(E) 1111\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111

Pick an answer.

(A)
11
(B)
19
(C)
101
(D)
111
(E)
1111
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Toolkit + CCSS Solution

Understand

Restated: A 6-digit number $Z$ has the special form where its first three digits repeat as its last three digits, like $247247$. The problem asks which of the five choices — $11, 19, 101, 111, 1111$ — must always be a factor of $Z$, no matter what those three repeated digits are.

Givens: $Z$ is a 6-digit positive integer; $Z$ has the form $abcabc$ — the first three digits ($abc$) are repeated as the last three digits ($abc$); Example given in the problem: $247247$; Answer choices: (A) $11$, (B) $19$, (C) $101$, (D) $111$, (E) $1111$

Unknowns: Which of the five choices is guaranteed to divide every number of the form $abcabc$

Understand

Restated: A 6-digit number $Z$ has the special form where its first three digits repeat as its last three digits, like $247247$. The problem asks which of the five choices — $11, 19, 101, 111, 1111$ — must always be a factor of $Z$, no matter what those three repeated digits are.

Givens: $Z$ is a 6-digit positive integer; $Z$ has the form $abcabc$ — the first three digits ($abc$) are repeated as the last three digits ($abc$); Example given in the problem: $247247$; Answer choices: (A) $11$, (B) $19$, (C) $101$, (D) $111$, (E) $1111$

Plan

Primary tool: #9 Solve an Easier Related Problem

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

Instead of jumping into algebra, we use Tool #9 (Easier Problem) by working with the concrete example $247247$ the problem hands us, plus one or two more $abcabc$ numbers. Tool #5 (Pattern) helps us notice that every $abcabc$ number is exactly $abc \times 1001$ — once we see that pattern, the question reduces to "which choices divide $1001$?". Finally Tool #3 (Eliminate Possibilities) checks the five choices against $1001$ to pick the winner. We deliberately avoid Tool #13 (Algebra) — the place-value pattern is easier to discover by playing with one example than by setting up variables first.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.NBT.B.6 Step 1
  • Start with the example $247247$ that the problem gives us.
  • Divide it by the repeated three-digit block $247$ to see what's left.
$$247247 \div 247 = 1001$$

💡 Dividing a 6-digit number by a 3-digit number is a Grade 4 long-division skill — no algebra needed.

#5 Look for a Pattern 4.OA.C.5 Step 2
  • Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coincidence.
  • Try $315315 \div 315$.
$$315315 \div 315 = 1001$$

💡 Computing two cases and seeing the same quotient $1001$ is the Grade 4 way of confirming a number pattern before trusting it.

#9 Solve an Easier Related Problem 4.NBT.A.2 Step 3
  • Explain WHY the quotient is always $1001$, using place value.
  • The first three digits $abc$ sit in the hundred-thousands, ten-thousands, and thousands places, which is $1000$ copies of $abc$.
  • The last three digits are just $1$ more copy of $abc$.
  • So $abcabc = abc \times 1000 + abc = abc \times 1001$.
$$\overline{abcabc} = \overline{abc} \times 1000 + \overline{abc} = \overline{abc} \times 1001$$

💡 Reading $abcabc$ by place value — the first $abc$ is worth $1000$ times the second $abc$ — is exactly the Grade 4 multi-digit place-value standard.

#5 Look for a Pattern 4.OA.B.4 Step 4
  • Since $Z = \overline{abc} \times 1001$ for every such number, any factor of $1001$ is automatically a factor of $Z$.
  • So the real question becomes: which of $11, 19, 101, 111, 1111$ divides $1001$?
  • Find the prime factorization of $1001$.
$$1001 = 7 \times 11 \times 13$$

💡 Finding factor pairs of a whole number (here, $1001$) is the Grade 4 factor-pair / prime-or-composite standard.

#3 Eliminate Possibilities 4.OA.B.4 Step 5
  • Check the five choices against $1001$ to eliminate.
  • $11$ is one of the prime factors — it divides $1001$.
  • The others do not: $1001 \div 19 \approx 52.7$ (not whole), $1001 \div 101 \approx 9.91$, $1001 \div 111 \approx 9.02$, $1001 \div 1111 < 1$.
  • Only $11$ survives.
$$1001 \div 11 = 91 \;\checkmark \qquad \text{others: not whole}$$

💡 Testing each choice as a divisor of $1001$ is the same Grade 4 factor-pair check applied to a multiple-choice elimination — only the true factor stays standing.

[1] #9 4.NBT.B.6 Start with the example $247247$ that the problem gives us. Divide it by the repe
[2] #5 4.OA.C.5 Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coinc
[3] #9 4.NBT.A.2 Explain WHY the quotient is always $1001$, using place value. The first three di
[4] #5 4.OA.B.4 Since $Z = \overline{abc} \times 1001$ for every such number, any factor of $100
[5] #3 4.OA.B.4 Check the five choices against $1001$ to eliminate. $11$ is one of the prime fac

Review

Reasonableness: Spot-check with two different $abcabc$ numbers. $247247 \div 11 = 22477$ (whole) and $123123 \div 11 = 11193$ (whole) — $11$ works for both. Now try a choice that shouldn't always work, like $19$: $123123 \div 19 \approx 6480.16$ is not whole, so $19$ is not a guaranteed factor. Only $11$ — sitting inside $1001 = 7 \times 11 \times 13$ — survives for every $abcabc$. Answer (A) $\boxed{11}$ holds up.

Alternative: Tool #13 (Convert to Algebra) gives a one-line proof: let $N = \overline{abc}$, then $Z = 1000N + N = 1001N$, so every factor of $1001 = 7 \times 11 \times 13$ divides $Z$. From the choices, only $11$ is among $\{7, 11, 13\}$. The algebra route is shorter, but the pattern-discovery route in our main path makes the $1001$ appear naturally instead of being announced.

CCSS standards used (min grade 4)

  • 4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing the concrete example $247247$ by $247$ to discover that the quotient is $1001$ — uses Grade 4 long division (the extra digits beyond four-digit dividends are handled by the same algorithm).)
  • 4.OA.C.5 Generate a number or shape pattern following a given rule (Confirming with a second example ($315315 \div 315 = 1001$) that every $abcabc$ number follows the same pattern before trusting the rule.)
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Reading the 6-digit number $abcabc$ by place value, recognizing the first $abc$ block is worth $1000$ times the second $abc$ block, which gives $abcabc = abc \times 1001$.)
  • 4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Factoring $1001 = 7 \times 11 \times 13$ and then testing each multiple-choice option ($11, 19, 101, 111, 1111$) as a possible divisor of $1001$.)

⭐ This AMC 8 problem only needs Grade 4 place value and factor pairs you already know — once you see that $abcabc = abc \times 1001$, it's just asking which choice divides $1001$!

⭐ This AMC 8 problem only needs Grade 4 place value and factor pairs you already know — once you see that $abcabc = abc \times 1001$, it's just asking which choice divides $1001$!

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