What is the largest palindrome that is smaller than 654321 and is a sum of nine consecutive integers?
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The first three digits must be the largest integer < 6 5 4 that is divisible by 9 . This number is 6 4 8 , and the solution is 6 4 8 8 4 6 .
The sum of 9 consecutive integers can be thought algebraically where 'x' is the original number, and (x+1), (x+2), (x+3)... are the following 8 consecutive numbers. Adding them together, we get (9x + 36).
Therefore, in order to satisfy this condition, we should be able to recreate the answer with (9x+36), where x is an integer. As 36 is already divisible by 9, this simplifies down to the fact that our answer must be divisible by 9.
Now, a quick way to find whether a number is divisible by 9 or not is to take its digital root. If the digital root is divisible by 9, the number itself is also divisible by 9.
First, let's assume the answer looks like '65xx56'. In this case, the digital root is 2x+22 In order for that to be divisible by 9, x=7 (2*7 + 22 = 36). However, 657756 > 654321.
The next biggest answer would be in the form of '64xx46'. Once again, checking the digital root, we find that it is 2x+20, and in order for the digital root to be divisible by 9, x=8 (2*8 + 20 = 36). As 648846 < 654321, that's our answer.
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Largest output= 6 4 8 8 4 6
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Since the palindrome is a sum of 9 consecutive integers, then it is of the form x +(x+1)+...+(x+8)= 9(x+4).Hence, it must be divisible by 9. We simply construct a palindrome less than and nearest to 654321 that is divisiblble by 9. The nearest is 648846.