2020 AIME I Problem 3

Question

A positive integer

N

has base-eleven representation

\underline{a}\,\underline{b}\,\underline{c}

and base-eight representation

\underline{1}\,\underline{b}\,\underline{c}\,\underline{a},

where

a,

b,

and

c

represent (not necessarily distinct) digits. Find the least such

N

expressed in base ten.

Answer

Answer: 621

Concepts:number base digits bounding to limit cases

Difficulty rating: 2110

Small hint:

Set the base-ten values equal:

121a + 11b + c

= 512 + 64b + 8c + a.

Big hint:

Simplify to

120a = 512 + 53b + 7c

with digits

a, b, c \le 7.

The right side forces

a \ge 5,

and smaller

a

means smaller

N.

Solution:

Equating the two representations in base ten gives

121a + 11b + c

= 512 + 64b + 8c + a,

which simplifies to

120a = 512 + 53b + 7c.

All of

a,

b,

c

are base-eight digits, so

0 \le a, b, c \le 7

(and

a \ge 1

since it leads the base-eleven representation).

The right side is at least

512,

so

a \ge 5.

Since

N = 121a + 11b + c

increases with

a,

try

a = 5:

then

53b + 7c = 88.

Here

b = 0

gives

7c = 88,

impossible, and

b \ge 2

overshoots, so

b = 1

and

7c = 35,

giving

c = 5.

Thus

N = 121 \cdot 5 + 11 + 5 = 621,

whose base-eight representation is

1155

and base-eleven representation is

515,

as required. The least such

N

is

621.