(100A+10B+C-100C-10B-A) is divisible by 7. Hence (99A-99C) is divisible by 7 or 99(A-C) is divisible by 7. The possibilities are A=9 and C=2 along with C=9 and A=2 or A=8 and C=1 along with A=1 and C=8. There are four numbers and nine possibilities for(A-C) turning into 0, i.e. 4+9=13 possibilities.
With (A-C)=0 we have one divisor (factor) and with (A-C)=2 we may have the following options: 9B2-2B9=693 has one factor 99. The same with every other number, e.g. 2B9-9B2=-693 has one factor -99. Hence total makes (4+1)=5 divisors (factors) out of which only
3 divisors will be different 0,99,-99
gopinathhyd wrote:1. If A, B and C are all positive integers. and three digit number ABC - CBA is divisible by 7. How many different numbers are possible divisors for 7.
Last edited by pemdas on Sun Mar 18, 2012 2:29 pm, edited 1 time in total.
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