Question 268466: A 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 11.
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! A 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 11
divisibility by 11 test: 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
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