+656 A palindrome is a number that is the same when read forwards and backwards, such as 43234. Find the number of 4-digit palindromes that are divisible by 3.
This question is very similar to other questions asked but I have checked all the web2.0calc and GMat club answers and they are all different questions, so please don't copy and paste your answer. Thanks!
A number is divisible by 3 if the sum of its digits is divisible by 3. Let the four-digit palindrome be abba, where a and b are digits. Then abba is divisible by 3 if and only if 2a+2b is divisible by 3. Since 2 is a factor of 3, this is equivalent to a+b being divisible by 3.
For each of the 9 choices for the digit b, there are exactly 3 choices for the digit a such that a+b is divisible by 3. Therefore, there are 9⋅3=27 4-digit palindromes that are divisible by 3.
There are 30 such palindromes as follows:
(1221, 1551, 1881, 2112, 2442, 2772, 3003, 3333, 3663, 3993, 4224, 4554, 4884, 5115, 5445, 5775, 6006, 6336, 6666, 6996, 7227, 7557, 7887, 8118, 8448, 8778, 9009, 9339, 9669, 9999) >>Total = 30 such palindromes.
The sum of the digits of each palindrome is a multiple of 3, thereby is divisible by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. So, a 4-digit palindrome is divisible by 3 if the sum of its first two digits and the sum of its last two digits are both divisible by 3.
Let the first two digits of the palindrome be ab and the last two digits be ba. Then, the sum of the digits is a+b+a+b=2(a+b). Since 2(a+b) is always even, it is divisible by 3 if and only if a+b is divisible by 3.
The possible values of a+b are 0, 3, 6, and 9. For each of these values, there are 9 choices for a and b, giving a total of 9×9=81 possible palindromes.
However, we need to remove the palindromes that are not divisible by 3. The only palindromes that are not divisible by 3 are those with a+b=0. There are 3 such palindromes: 1111, 2222, and 3333.
Therefore, there are 81−3=78 4-digit palindromes that are divisible by 3.
