2018 AMC 12A 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 algebraic manipulation

Difficulty rating: 2650

Small hint:

Write

A_n = a \cdot \tfrac{10^n - 1}{9},

and likewise

B_n

and

C_n = c \cdot \tfrac{10^{2n} - 1}{9},

then substitute into

C_n - B_n = A_n^2

Big hint:

Requiring it for two values of

n

forces the coefficient of

10^n

to vanish:

9c = a^2

and

9b - 9c = a^2

Solution:

Using

A_n = a \cdot \tfrac{10^n - 1}{9},

B_n = b \cdot \tfrac{10^n - 1}{9},

and

C_n = c \cdot \tfrac{10^{2n} - 1}{9},

the equation

C_n - B_n = A_n^2

becomes, after dividing by

10^n - 1

and clearing fractions,

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

For this to hold at two different

n,

the coefficient of

10^n

must be zero, so

9c = a^2

and hence

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

Then

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

and

b = 2c.

So

a \in \{3, 6, 9\}

with

c \in \{1, 4, 9\}

and

b \in \{2, 8, 18\};

the case

b = 18

is not a digit. The valid triples are

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

and

(6, 8, 4),

and indeed

4444 - 88 = 4356 = 66^2.

The greater digit sum is

6 + 8 + 4 = 18.

Thus, the correct answer is D.

2018 AMC 12A Problem 25

Answer: D

Difficulty rating: 2650

Small hint:

Big hint:

Solution:

Problem 25 in other years