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....