Sum of Palindromes
What is the sum of all 4-digit palindromes ?
Clarification : Palindromes are the number which are read the same both forward and backward. For eg, 343 and 7887 are palindromes while 123 and 998 are not.
The answer is 495000.
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4 solutions
Sir, I was trying to answer your old logic level 4 question on Google but could not get submitted. And I have got a surely right answer. Please reply me as soon as possible. I will humbly accept your advice
For each nonzero digit x, a 4-digit palindrome whose first digit is x are x00x,x11x,x22x,...x99x. Adding these numbers using each digit's place value, we will have x 0 0 x + x 1 1 x + . . . + x 9 9 x = 1 0 3 ( 1 0 x ) + 1 0 x + 1 0 2 ( 1 + 2 + . . . + 9 ) + 1 0 ( 1 + 2 + . . . + 9 ) = 1 0 4 x + 1 0 x + 1 0 0 ( 4 5 ) + 1 0 ( 4 5 ) = 1 0 4 x + 1 0 x + 4 4 5 5 0
Ranging x from 1 to 9, then the sum of all 4-digit palindrome is given by x = 1 ∑ 9 1 0 3 ( 1 0 x ) + 1 0 x + 1 0 2 ( 1 + 2 + . . + 9 ) + 1 0 ( 1 + 2 + . . . + 9 ) = x = 1 ∑ 9 1 0 4 x + 1 0 x + 4 4 5 5 0 = 4 9 5 0 0 0 .
Write all of them out, then add them up.
Adding 90 numbers is a pretty tedious task. You can use the fact that the numbers are periodic. Check Jerry's solution for a better way.
Simpler... very very easy problem :(
First, we shall find the number of 4 digit palindromes, which is 90(9 numbers for first and last digit, 10 numbers for second and third digit). Now, we can assume that each digit occurs as frequently as every other digit. Hence, for the thousands and ones digit, the "average" digit should be 5 (from 1 to 9). For the hundreds and tens digit, the "average" digit should be 4.5 (from 0 to 9). Thus, we get the average of all these palindromes, which is 5500. Then, multiplying by 90, we get the answer 495000.