What term is the smallest five-digit palindrome in the arithmetice sequence 2 7 12 17?

Answer 1

20002.

The required answer is a 5 digit number which reads the same forwards and backwards, so it is of the form abcba where a, b & c are individual digits, that is the first and last digits are the same, the second and fourth digits as the same, and the first, second and third digits may, but not necessarily, be different; the problem comes down to finding a, b & c so that abcba is as small as possible.

The arithmetical sequence runs 2, 7, 12, 7,...; the difference between the terms is 5, so the nth term is:

an = 5n - 3

However, the first thing to notice is that the last digit alternates between 2 & 7, so the first and last digit, a, must be 2 or 7; all five digit numbers starting with 2 are smaller than all five digit numbers starting with 7 so a=2 will most likely give the smallest five digit palindrome - so now b & c needs to be found to make 2bcb2 as small as possible.

The obvious starting point would be to try with b and c as small as possibly, namely zero (b=c=0) giving 20002 and see if that is part of the sequence - if it is, it must fit with the nth term given above, namely for some integer n:

20002 = 5n - 3

=> 5n = 20005

=> n = 4001

As n = 4001 is an integer, 20002 is part of the sequence.

The previous number in the sequence is 19997, so all previous five digit numbers cannot be palindromes (since as the last digit is either 2 or 7, the first digit must also be 2 or 7) so 20002 must be the smallest 5 digit palindrome.

Answer 2

It is 20002.