The positive five-digit integer is in the form AB, CBA: where A,B and C are each distinct digits. what is the greatest possible value of AB,CBA that is divisible by eleven?

Question

Please help with this problem

Answer 1

Divisibility by 11 is tested by taking the difference of the sums of alternate digits: A+C+A-2B=11n, where n is an integer. Put A=9 and B=8: 2A+C-2B=18+C-16=11n. C+2=11n. If n=1, C=9. The number is therefore 98,989. But C has to be distinct. Reduce B to 7: 18+C-14=11; C=11-4=7=B, so again C is not distinct. But a pattern is emerging. Reduce B to 6: 18+C-12=11 and C=5, which is distinct, and the number is 96,569.

Answer 2

The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11
example: 34871903
3+8+1+0=12
4+7+9+3=23
23-12=11
Is divisible by 11
so we want (b + b) - (a + c + a) to be divible by 11
2b - (2a + c) to be divisible by 11
ab,cba (using 7 and 8 and 9 since they biggest)
78 987 --> 2*8 - (2*7 + 9) = 16 - (14 + 9) = 16 - 23 = -7 NO
87 978 --> 2*7 - (2*8 + 9) = 14 - (16 + 9) = 14 - 25 = -11 YES
79 897 --> 2*9 - (2*7 + 8) = 18 - (14 + 8) = 18 - 22 = -4 NO
97 879 --> 2*7 - (2*9 + 8) = 14 - (18 + 8) = 14 - 26 = -12 NO
89 798 --> 2*9 - (2*8 + 7) = 18 - (16 + 7) = 18 - 23 = -5 NO
98 789 --> 2*8 - (2*9 + 7) = 16 - (18 + 7) = 16 - 25 = -9 NO
so biggest 5 digit number is 87,978

Positive Integers