AMC 10 · 2007 · #22
Grade 6 number-theoryA finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let S be the sum of all the terms in the sequence. What is the largest prime factor that always divides S?
Pick an answer.
AMC 10 2007 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 row of three-digit numbers is arranged in a loop: the tens and units digits of each number become the hundreds and tens digits of the next number, and the loop closes so the last number's last two digits feed back into the first number. Add up every number in the loop to get $S$. Find the largest prime that must divide $S$ for every loop that follows this rule.
Givens: Every term is a three-digit integer; The tens and units digits of each term equal, in order, the hundreds and tens digits of the next term; The rule wraps around: the tens and units digits of the last term equal the hundreds and tens digits of the first term; Example loop: $247,\ 475,\ 756,\ \dots,\ 824$; $S$ is the sum of all the terms; Answer choices: (A) $3$, (B) $7$, (C) $13$, (D) $37$, (E) $43$
Unknowns: The largest prime number that divides $S$ no matter which valid loop is chosen
Understand
Restated: A row of three-digit numbers is arranged in a loop: the tens and units digits of each number become the hundreds and tens digits of the next number, and the loop closes so the last number's last two digits feed back into the first number. Add up every number in the loop to get $S$. Find the largest prime that must divide $S$ for every loop that follows this rule.
Givens: Every term is a three-digit integer; The tens and units digits of each term equal, in order, the hundreds and tens digits of the next term; The rule wraps around: the tens and units digits of the last term equal the hundreds and tens digits of the first term; Example loop: $247,\ 475,\ 756,\ \dots,\ 824$; $S$ is the sum of all the terms; Answer choices: (A) $3$, (B) $7$, (C) $13$, (D) $37$, (E) $43$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #5 Look for a Pattern, #15 Organize Information in More Ways, #3 Eliminate Possibilities
The overlap rule means each digit is shared by three consecutive terms, so the loop is really one ring of digits $d_1, d_2, \dots, d_n$ written around a circle. Tool #4 (Introduce a Variable) names those digits and writes each term as $100 d_k + 10 d_{k+1} + d_{k+2}$. Tool #5 (Look for a Pattern) spots that every digit lands in the hundreds place once, the tens place once, and the units place once as you slide around the ring. Tool #15 (Organize Information in More Ways) adds the terms column by column instead of term by term, which collapses the sum to $111$ times the digit total. Then Tool #3 (Eliminate Possibilities) factors $111$ and uses the answer choices plus one tiny loop to pin down the largest prime that is forced every time.
Execute — Answer: D
6.EE.A.2 Step 1 Name the ring of digits
- Because the last two digits of each term are the first two digits of the next, consecutive terms share their digits.
- Follow one digit at a time around the whole loop and call the digits $d_1, d_2, \dots, d_n$ arranged in a circle, where $n$ is the number of terms.
- Then the $k$-th term has hundreds digit $d_k$, tens digit $d_{k+1}$, and units digit $d_{k+2}$, with the subscripts wrapping around ($d_{n+1} = d_1$, and so on).
- So each term is $100 d_k + 10 d_{k+1} + d_{k+2}$.
💡 The two-digit overlap glues the terms into one circular string of digits, so a single letter per digit describes the entire loop.
4.NBT.A.2 Step 2 Each digit visits every place once
- Write a term in expanded place-value form: $247 = 100\cdot 2 + 10\cdot 4 + 1\cdot 7$.
- Now track one digit, say $d_k$, around the ring.
- It is the hundreds digit of term $k$, the tens digit of the term before it (term $k-1$), and the units digit of the term two before it (term $k-2$).
- Since the loop is closed, this is true for every digit.
- So as you total all the terms, each digit $d_k$ gets counted exactly once with weight $100$, once with weight $10$, and once with weight $1$.
💡 Sliding one step around the ring shifts a digit from the hundreds place to the tens place to the units place, so it collects each place value exactly once.
6.EE.A.3 Step 3 Add by place value, not term by term
- Instead of adding term by term, add up each place-value column across all terms.
- The hundreds column contributes $100(d_1 + d_2 + \dots + d_n)$, the tens column contributes $10(d_1 + \dots + d_n)$, and the units column contributes $1(d_1 + \dots + d_n)$, because every digit shows up once in each column.
- Let $T = d_1 + d_2 + \dots + d_n$ be the total of all the digits.
- Then $S = 100T + 10T + T = 111\,T$.
💡 Regrouping the sum by columns pulls the shared digit total $T$ out front, leaving the fixed factor $100+10+1 = 111$.
4.OA.B.4 Step 4 Factor 111 and pin the largest forced prime
- Since $S = 111\,T$ for every valid loop, whatever divides $111$ always divides $S$.
- Factor it: $111 = 3 \times 37$, and both $3$ and $37$ are prime.
- So $3$ and $37$ divide $S$ every time, and $37$ is the larger of the two.
- Could a bigger prime be forced?
- No: the single-term loop $111$ (digits $1,1,1$) is legal and gives $S = 111$, whose prime factors are only $3$ and $37$.
- So no prime above $37$ — and in particular none of $7$, $13$, or $43$ from the choices — can divide every $S$.
- The largest prime that always divides $S$ is $37$, choice (D).
💡 Every loop's sum is a multiple of $111 = 3\times 37$, so $37$ is guaranteed, and the bare loop $111$ proves no larger prime can be.
6.EE.A.2 Because the last two digits of each term are the first two digits of the next, c 4.NBT.A.2 Write a term in expanded place-value form: $247 = 100\cdot 2 + 10\cdot 4 + 1\cdo 6.EE.A.3 Instead of adding term by term, add up each place-value column across all terms. 4.OA.B.4 Since $S = 111\,T$ for every valid loop, whatever divides $111$ always divides $ Review
Reasonableness: Test the given example digits. The ring $2,4,7,5,6,\dots$ makes terms $247, 475, 756, \dots$; whatever the full ring is, the claim says $S$ must be $111$ times the digit total. Check a complete small loop directly: $247 + 475 + 752$ uses the ring $2,4,7,5$ closing back, and $247+475+752+524 = 1998 = 111 \times 18$, and $2+4+7+5 = 18$ — the $111\,T$ rule holds. Both $3$ and $37$ divide $1998$. The trap answers are the other primes near $37$: $7$, $13$, and $43$ each divide some loops but not the plain loop $111$, so none is guaranteed; and $3$ works always but is smaller than $37$. So $37$ being the largest forced prime is consistent.
Alternative: Skip naming a general ring and just argue by counting roles. In any closed loop, list every digit that appears anywhere. Each such digit sits in the hundreds place of exactly one term, the tens place of exactly one term, and the units place of exactly one term. So its total contribution to $S$ is $100 + 10 + 1 = 111$ times its value. Summing over all digits gives $S = 111 \times (\text{sum of digits})$ immediately, and factoring $111 = 3 \times 37$ gives $37$.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the ring of digits $d_1,\dots,d_n$ and writing each term as $100 d_k + 10 d_{k+1} + d_{k+2}$.)4.NBT.A.2Read and write multi-digit whole numbers using base-ten numerals and expanded form (Expanding each three-digit term into $100\cdot(\text{hundreds}) + 10\cdot(\text{tens}) + 1\cdot(\text{units})$ to track each digit's place.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Regrouping the sum by place-value columns into $100T + 10T + T = 111\,T$.)4.OA.B.4Find all factor pairs for a whole number and determine whether it is prime or composite (Factoring $111 = 3 \times 37$ and identifying $37$ as the largest prime that always divides $S$.)
⭐ Because the loop makes every digit land in the hundreds, tens, and units place exactly once, the whole sum is always $111 = 3 \times 37$ times the digit total, so $37$ is the biggest prime guaranteed to divide it.
⭐ Because the loop makes every digit land in the hundreds, tens, and units place exactly once, the whole sum is always $111 = 3 \times 37$ times the digit total, so $37$ is the biggest prime guaranteed to divide it.
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