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?

Question 13

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?

Solution

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?

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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?

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