AMC 10 · 2018 · #25
Grade 8 arithmeticPick an answer.
AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Write each repdigit with a repunit
Each repdigit is (its digit)·(10^k-1)/9, so A_n=a·(10ⁿ-1)/9, B_n=b·(10ⁿ-1)/9, C_n=c·(10²ⁿ-1)/9.
Every repdigit is just its digit times a string of ones, and a string of ones is a power of ten minus one over nine.
8.EE.A.1Convert To AlgebraLet x=10ⁿ
Set x=10ⁿ (so 10²ⁿ=x²); the equation C_n-B_n=A_n² becomes c(x²-1)/9-b(x-1)/9=a²(x-1)²/81, one equation in x.
Renaming 10ⁿ as x turns a statement about digit counts into ordinary algebra.
8.EE.A.1Use Matrix LogicCancel the common factor x-1
Since x²-1=(x-1)(x+1) and x=10ⁿ≥10, divide out x-1 (multiply by 81, divide by 9(x-1)), leaving 9c(x+1)-9b=a²(x-1).
The shared factor x-1 divides out cleanly because x=10ⁿ is never 1.
7.EE.A.1Convert To AlgebraCollect into a linear equation in x
Expand and gather: 9cx+9c-9b=a²x-a² becomes (9c-a²)x+(a²+9c-9b)=0, linear (first-degree) in x.
Lining the equation up as (slope)· x+(constant)=0 exposes exactly two numbers that control everything.
8.EE.C.7Convert To AlgebraTwo values of n force both coefficients to zero
A linear Mx+K=0 has ≤1 root unless M=K=0; needing two x=10ⁿ makes it an identity, so 9c-a²=0 and a²+9c-9b=0.
A straight line crosses zero only once, so hitting zero twice means the line is flat at zero.
8.F.A.1Use Matrix LogicSolve the system for the digit relations
The first equation gives a²=9c; substitute into the second: 18c=9b, so a²=9c and b=2c — choosing c fixes a and b.
The two coefficient equations pin a and b directly to c.
8.EE.C.8Use Matrix LogicUse digit limits to find legal triples
c=a²/9 forces a∈{3,6,9}: a=3→(c,b)=(1,2); a=6→(c,b)=(4,8); a=9 needs b=18, not a digit.
a² must be nine times a digit, so a has to be divisible by 3, leaving only three cases to check.
7.EE.B.4Eliminate PossibilitiesMaximize the sum
(3,2,1) gives sum 6; (6,8,4) gives 6+8+4=18, the greatest — choice (D).
Among the surviving cases, the largest digits give the largest sum.
8.EE.C.7Evaluate Finite DifferencesTurn 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