A palindrome is a number that reads the same forward and bac

Hi experts Bunuel IanStewart

This question took me about three minutes in my practice exam, but I still got it incorrectly.
I was not familiar with palindrome, and I did not see that the number of combinations could be obtained by 3*3*3. Instead, I tried to count the possibilities manually. (The answer choices range between 12 to 27, which is not too many to count. It is not the most efficient way, but I had no other ways then.) I failed because I only thought of three types of arrangements, AAAAA, ABCBA and ABBBA, ignoring other two, AABAA and ABABA.

I checked the solutions in this thread (they are smart) and practiced other two similar palindrome questions.
I hope to confirm two issues:

1. When a palindrome has an even number of digits, we only need to care about the first half. When the palindrome has an odd number of digits, we care about the first half and the middle one.

For example, for the 4-digit palindrome XYZW, we only chose numbers for X and Y, since Z and W just copy the preceding digits respectively.
For the 5-digit palindrome ABCDE, we only chose numbers for A,B and C, letting D and E copy A and B.

2. These palindrome questions may vary in constraints, such as "the number must be odd," "no digits can be repeated," "the digits could only be 1, 2 and 3." We just tackle each questions differently.


Thank you.

You are correct about both points!!

I wouldn't suggest thinking of palindrome counting as a 'question type' (and thus learning some method for this precise situation). This might be the only official question that will ever test palindromic numbers. It's better to think of this question as just a standard counting problem where earlier choices constrain later choices. When we pick our first digit here, because the number is a palindrome, that uniquely determines the final digit, and when we pick the second digit that determines the fourth digit. That's a situation that could be tested in any number of ways -- for example, you could be asked how many six digit numbers there are where all of the odd-positioned digits are the same as each other, and all of the even-positioned digits are the same as each other, numbers like 373737. That's not a palindrome question, but it's testing the same idea that this palindrome question tests.

All of that said, the answer to both of your questions is 'yes'.

IanStewart
Thank you for your explanations!

Your final question is really interesting.
Is the answer 9*9=81 if the digits in the odd and even positions could not be repeated?
If the digits could be repeated, the answer would be 9*10=90.

Yes, if I had taken more care to word the question well, it would have specified whether the odd and even placed digits could be equal, but both of your answers are correct.

I understand the correction, but in the statement it says "If 5-digit palindromes are formed using one or more of the digits".
So for me, each of the integers 1, 2 or 3 must be part of the palyndrome at least once, which makes versions 12221 or 22122 where the 3 doesn't appear impossible, for example - no one agrees ?

The phrase "using one or more of the digits" does not mean each digit must appear. It just means digits must be chosen from the set {1, 2, 3}. So palindromes like 12221 or 11111 are valid.