This is a 4th grade question. A palindrome is an integer number that does not change when read backwards. E.g. 123321 is a 6 digit palindrome. How many 9-digit palindromes are there that use only the digits 9 and 0?
I have created a table and typing the possible numbers, but this takes more time than needed. Kindly advise the best approach to solve this problem so that a 4th grader can understand that.
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If you want to get a $9$-digit number, you can't use $0$ at the first place. Therefore, the first digit must be $9$, and since the number is palindrome, the last digit must be $9$ too. Now you are limited to make $7$ digit numbers($0$ can be first too) which are symmetrical. You can choose the first $4$ digits howewer you want, these are $2*2*2*2$ options(in every digit, you can choose $0$ or $9$), and you must choose the last $3$ digits to make it a palindrome, but these choices are depending on what you chose at the first $4$ options.
Your answer is therefore $16$. Hope this is understandable for your studies.
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By Rule of Product, in questions like these we set up a sequence of questions with each having a set number of possible responses such that:
As we are asking for palindrome nine-digit integers, we set up the following picture to help remind us how to approach the problem.
$\underline{~a~}~\underline{~b~}~\underline{~c~}~\underline{~d~}~\underline{~e~}~\underline{~d~}~\underline{~c~}~\underline{~b~}~\underline{~a~}$
We set up the following sequence of questions:
Multiplying the number of options at each step, we have a total of $1\cdot 2\cdot 2\cdot 2\cdot 2\cdot 1\cdots 1 = 2^4=16$ possible palindromic 9-digit integers using only digits zero and nine.
The digit cannot start with 0 as it becomes a 8 digit no so the first no should be 9 so the last digit. Let the remaining 7 nos be XYZPZYX. Here P can take values of 0 or 9 so two cases for P.Similarly for X,Y,Z.So total nos are 2*2*2*2(for values of X,Y,Z,P)=16