Problem:
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=444eight .
Solution:
Answer (585):
Let N be a three-digit palindrome in base ten. Then there are digits a and b such that N=100a+10b+a=101a+10b. Suppose that N has four digits when written in base eight. Because 2⋅83=1024>1000, in base eight N must start with the digit 1 . Hence N=1cc1eight =512+64c+8c+1=513+72c=101a+10b. Note that the units digit of 101a+10b equals a. Because 513+72c is odd, a must also be odd. Consider the possible values of a in decreasing order.
- If a=9, then 513+72c=909+10b. Hence 2c≡6(mod10), and thus c=3 or c=8. However, if c=3, then 513+72c<909, and c cannot be 8 as c is a digit in base eight.
- If a=7, the only possible values of c are 2 and 7. But if c=2, then 513+72c<707, and if c=7, then 513+72c>1000. Hence neither case is possible.
- If a=5, the only possible values of c are 1 and 6. If c=6, then b=44, which is not allowed. If c=1, then b=8 is a solution, leading to N=505+80=585=1111eight .
It is not necessary to consider the case when N has only 3 digits when written in base eight because any such palindrome would be less than 585 . Hence the requested palindrome is 585 .
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions