2023 AIME II Problem 2

Question

Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than

1000

that is a palindrome both when written in base ten and when written in base eight, such as

292 = 444_8.

Answer

Any base-eight palindrome between

512

and

999

has four base-eight digits and looks like

\overline{1bb1}_8 = 513 + 72b

Three-digit base-eight numbers are at most

511,

so just test

b = 6, 5, 4, \ldots

for a base-ten palindrome

A four-digit base-eight number lies between

512

and

4095,

so a base-eight palindrome less than

1000

with four digits must have leading (and trailing) digit

1:

it has the form

\overline{1bb1}_8 = 512 + 64b

{}+ 8b + 1

= 513 + 72b.

Keeping this below

1000

requires

b \le 6,

giving the candidates

513, 585, 657, 729, 801, 873, 945.

Checking from the top, the only one of these that is also a palindrome in base ten is

585 = 1111_8.

Every base-eight palindrome with at most three digits is at most

777_8 = 511 \lt 585,

so the answer is

585.

Problem 2 in other years