Explanation:Any 3-digit number \(xyz\) can be written as \(100x + 10y + z\)
\(abc = 100a + 10b + c\)
\(cba = 100c + 10b + a\)
So, \(D = 99a - 99c = 99(a - c)\)
If \(D\) is a multiple of \(7\) then \((a - c)\) must be \(7\)
Let us look for cases for which \((a - c) = 7\);
Case I: a = 9 and c = 2Case II: a = 8 and c = 1Case III: a = 7 and c = 0Now, we can have two feasible cases and \(b\) can take 10 values
So, Number of +ve values of \(abc\) which are divisible by \(7 = (2)(10) = 20\)
Col. A: 20
Col. B: 15
Hence, option A
NOTE: Case III is invalid as c = 0, \(cba\) becomes a 2-digit integer
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/