Silver89 wrote:

k=100x + 10y + z

99+k= 100z + 10y + x

thus 99+ 100x + 10y + z = 100z + 10y + x

99= 99z - 99x

so z = x+1

z must be no more than 8 , if it is 9 for example so x=10 and that impossible

so we have 8 possibility in each 100 and we have 10 hundreds in 1000

so 8*10= 80

you are welcome man

please consider giving kudos

I solved it the same way;

If k=xyz

x+1=z

The numbers will be:

1@2 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10

2@3 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10

3@4 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10

4@5

5@6

6@7

7@8

8@9 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10

10*8=80

Ans: "D"

_________________

~fluke

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