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The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?
  • a)
    100<A<299
  • b)
    106<A<305
  • c)
    112<A<311
  • d)
    118<A<317
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The digits of a three-digit number A are written in the reverse order ...
 Let A = 100x + 10y + z  and B = 100z + 10y + x .According to given condition B - A = 99(z - x) As (B - A) is divisible by 7 . So clearly  (z - x) should be  divisible by 7.  z and x can have values 8,1 or 9,2 , such that 8-2=9 2=7 and  y can have  value from 0 to 9.
So Lowest possible value of A lowest x,y and z which is  is 108 and the highest possible value of A is 299.
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The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?a)100<A<299b)106<A<305c)112<A<311d)118<A<317Correct answer is option 'B'. Can you explain this answer?
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