AMC 8 · 2017 · #7
Grade 4 number-theoryProblem
Let Z be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of Z?
Pick an answer.
AMC 8 2017 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: 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
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.
💡 Dividing a 6-digit number by a 3-digit number is a Grade 4 long-division skill — no algebra needed.
4.OA.C.5 Step 2 - Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coincidence.
- Try $315315 \div 315$.
💡 Computing two cases and seeing the same quotient $1001$ is the Grade 4 way of confirming a number pattern before trusting it.
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 $abcabc = abc \times 1000 + abc = abc \times 1001$.
💡 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.
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 $1001$.
💡 Finding factor pairs of a whole number (here, $1001$) is the Grade 4 factor-pair / prime-or-composite standard.
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.
💡 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.
4.NBT.B.6 Start with the example $247247$ that the problem gives us. Divide it by the repe 4.OA.C.5 Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coinc 4.NBT.A.2 Explain WHY the quotient is always $1001$, using place value. The first three di 4.OA.B.4 Since $Z = \overline{abc} \times 1001$ for every such number, any factor of $100 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.6Find 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.5Generate 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.2Read 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.4Find 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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