AMC 10 · 2018 · #25

Grade 8 number-theory
digit-decompositionsequences-geometricplace-valuesystems-of-equations convert-to-algebraextremal-constructioncasework ↑ Prerequisites: digit-decompositionsequences-geometric
📏 Long solution 💡 4 insights
Problem
Pick three nonzero digits. Build three repeated-digit numbers: the first repeats one digit n times, the second repeats another digit n times, and the third repeats a third digit two n times. We want the third minus the second to equal the square of the first for at least two different values of n. Among all digit triples that allow this, find the greatest possible sum of the three digits and match it to a choice.

Pick an answer.

(A)
12
(B)
14
(C)
16
(D)
18
(E)
20
How to solve
Strategy Convert to Algebra

Tool #13 (Convert to Algebra): a repdigit of k equal digits d equals d·(10^k-1)/9, so the word problem about strings of digits becomes one clean equation. Tool #4 (Introduce a Variable): set x=10ⁿ so the equation turns into a polynomial in x; 'two values of n' becomes 'two values of x'. The key leverage is that a degree-one (linear) equation in x can be satisfied by two different x only if it is an identity, which forces its coefficients to be zero. Tool #14 (Extreme Principle): once the digit relations are pinned down, push a, b, c as large as the digit rules allow to maximize the sum. Tool #3 (Eliminate Possibilities): test the few legal values of a and throw out the one whose b exceeds 9.

1STEP 1

Write each repdigit with a repunit

Write each one with a repunit.

A_n=a·(10ⁿ-1)/9, B_n=b·(10ⁿ-1)/9, C_n=c·(10²ⁿ-1)/9
2STEP 2

Let x=10ⁿ

Let one letter be ten to the n.

(c (x²-1))/9-(b (x-1))/9=(a² (x-1)²)/81, x=10ⁿ
3STEP 3

Cancel the common factor x-1

Cancel the common factor.

9c (x+1)-9b=a² (x-1)
4STEP 4

Collect into a linear equation in x

What remains is linear.

(9c-a²) x+(a²+9c-9b)=0
5STEP 5

Two values of n force both coefficients to zero

Two values of n force both coefficients to zero.

9c-a²=0 and a²+9c-9b=0
6STEP 6

Solve the system for the digit relations

The system gives relations between the digits.

a²=9c, b=2c
7STEP 7

Use digit limits to find legal triples

The single-digit limit leaves two triples.

a=3→(c,b)=(1,2); a=6→(c,b)=(4,8); a=9→ b=18 (invalid)
8STEP 8

Maximize the sum

Taking the bigger one, the sum is 18.

a+b+c=6+8+4=18 → (D)
Answer
18
Check (a,b,c)=(6,8,4) at n=1: A₁=6, B₁=8, C₁=44. Then C₁-B₁=44-8=36=6²=A₁². At n=2: A₂=66, B₂=88, C₂=4444. Then C₂-B₂=4444-88=4356=66²=A₂². It works for two values of n, exactly as required, and a+b+c=18 matches choice (D). The rejected triple a=9 would need b=18, which is not a digit, so 18 is genuinely the ceiling.
💡Key takeaway

Turn each block of repeated digits into (digit)×(10^k-1)/9; the equation becomes linear in 10ⁿ, and since a line hits zero only once, holding for two n forces a²=9c and b=2c, whose biggest legal digits 6,8,4 add to 18, choice (D).

  • Write each repdigit with a repunit
  • Let x=10ⁿ
  • Cancel the common factor x-1
  • Collect into a linear equation in x
  • Two values of n force both coefficients to zero
  • Solve the system for the digit relations
  • Use digit limits to find legal triples
  • Maximize the sum