2018 AMC 10A Problem 25

Question

For a positive integer

n

and nonzero digits

a,

b,

and

c,

let

A_n

be the

n

-digit integer each of whose digits is equal to

a

; let

B_n

be the

n

-digit integer each of whose digits is equal to

b

; and let

C_n

be the

2n

-digit (not

n

-digit) integer each of whose digits is equal to

c.

What is the greatest possible value of

a + b + c

for which there are at least two values of

n

such that

C_n - B_n = A_n^2?

12

14

16

18

20

Answer: D

Concepts:digits geometric sequence algebraic manipulation

Difficulty rating: 2390

Small hint:

Write the repeated-digit numbers as geometric sums

Big hint:

Having two different

n

values forces the linear equation in

10^n

to be identically true

Solution:

The repeated-digit numbers can be written as

\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9}. \end{aligned}

Substituting these expressions into

C_n-B_n=A_n^2

and dividing by

\frac{10^n-1}{9}

gives

c(10^n+1)-b=a^2\dfrac{10^n-1}{9}.

Rearranging yields

(9c-a^2)10^n=9b-9c-a^2.

If this holds for two different values of

n,

subtracting the two equations shows that

9c-a^2=0.

The displayed equation then also forces

9b-9c-a^2=0.

Hence

c=\frac{a^2}{9}

and

b=\frac{2a^2}{9}.

Because

a,b,c

are nonzero digits, the candidates are

(a,b,c)=(3,2,1),

(6,8,4),

and

(9,18,9).

The last triple is invalid because

18

is not a digit. The greatest valid sum is

6+8+4=18.

Thus, D is the correct answer.

2018 AMC 10A Problem 25

Answer: D

Difficulty rating: 2390

Small hint:

Big hint:

Solution:

Problem 25 in other years