Let abc and dcb represent three-digit positive integers.

Question

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

Bunuel · Accepted Answer

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.

EMPOWERgmatRichC · Answer

Hi anik19890,

To start, you have to look at the UNITS digits:

We know that C+B = a number that ends in 8...
So C+B = 8 OR
C+B = 18

Now look at the TENS digits:
Here, we ALSO have B+C....but the total ends in 9. How can THAT happen (when the total ended in 8 with the UNITs digit)?

IF, in the UNITS digit, C+B = 8, then in the TENS digit, B+C should also equal 8. Since that does NOT happen, C+B CANNOT = 8 (it must equal 18).

So, there must be a 'carryover' of 1 from the UNITS to the TENS.....and there must ALSO be a 'carryover' of 1 from the TENS to the HUNDREDS.

Thus, A+D+1 = 5

GMAT assassins aren't born, they're made,
Rich

gayam · Answer

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....

Lucky2783 · Answer

Answer :D . 
c+b gives a unit digit of 8 
b+c gives a unit digit of 9

clearly c,b=9 i.e. a+d=4

anik19890 · Answer

how do you conclude that a carry over 1 from units to hundreds since a and d is different , it could be 3+2 or 1+4

mvictor · Answer

ok, so we have
abc
dcb
----
598.

we see that b+c must be 18, otherwise we have twice b+c with different results. since the max value for digits could be 9, and since 18 is the maximum value of 2 digits, it must be the case that b and c are both 9.
we have
a99
d99
----
598.

ok, so 5 = a+d+1. since we have to carry 1 from 9+9 in the tens digit.
subtract 1:
4=a+d.
a=4-d.

we have only one answer choice that satisfies this condition.

SeregaP · Answer

abc
dcb
598

notie that though b, c is both in units and tens, and that the sum is different=>their sum is over 10=>+1 for the hundreds=>4-d is a
E

rishabhshreyas14 · Answer

a=d also does since both a and d = 2.

Archit3110 · Answer

given abc+dcb=598 so units and tens place have to be same i.e c=b=9 so 1 gets carried to a+d+1 = 5 so a+d=4 so a = 4-d option D

PandCduo · Answer

abc= 598-dcb
a= 598/bc - d

Cancel 1 and 2

Now 598/3 will be a decimal and not absolute
Also 598/5 will not be absolute

Option D is correct

Posted from my mobile device

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Let abc and dcb represent three-digit positive integers.

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