Bunuel wrote:

gmat6nplus1 wrote:

Let abc and dcb represent three-digit positive integers. If abc+dcb=598, then which of the following must be equivalent to a?

A. d-1

B. d

C. 3-d

D. 4-d

E. 5-d

\(abc\)

+

\(dcb\)

_____

\(598\)

Notice that for the units and tens digit we have the same sum: \(c+b\). But the result is 8 for the units digit and 9 for tens digit. This implies that there is a carry over 1 from units to tens, thus there is a carry over 1 from units to hundreds. Therefore \(a+d=4\) --> \(a=4-d\).

Answer: D.

Hi Bunuel,

I have followed the following method to solve the problem

100(a+d)+11(b+C) = 598

a+d can be anything like 5 or 4 or 3 etc not sure what it is..

if we leave out the hunderds digit and concentrate on the rest of the number... it should be a multiple of 11...

98 is not a multiple of 11... multiples of 11 are 11, 22, 33, 44... so on 198

598 - 198 = 400

so the hundreds digit is 4

a+d=4

a=4-d

let me know if this approach is good....