gmat_winter wrote:Given that R is a positive three-digit integer, what is the hundreds digit of R?
(1) The hundreds digit of 3R is 8.
(2) (R + 1) results in a number with the hundreds digit of 9.
OAC
There are two ways that the hundreds digit of 3R could be 8. One would involve a situation in which multiplying the hundreds digit of R by 3 would create a number ending in 8. For instance if the hundreds digit of R were 6, then as long as there were nothing to add from the multiplication of the tens and units digits, multiplying that digit by 3 would create a number 18, and the hundreds digit would be 8. R = 600 works this way.
The other way Statement 1 would work would involve creating a hundreds digit of 8 via the combination of multiplying the hundreds digit of R by 3 and adding something from the previous multiplication of the tens and units digits. For instance if R were 280, 3R would be 840.
So there are two different ways to get a 3R with a hundreds digit of 8, and those two ways involve different hundreds digits for R. So Statement 1 is insufficient.
Statement 2 would be satisfied by any R to which we can add 1 and stay in the 900-999 range. So we can subtract one from 900 and get R = 899, which has a hundreds digit of 8 and alternatively we could subtract 1 from 999 to get R = 998, which has a hundreds digit of 9. So we have two possibilities for hundreds digits for R, and Statement 2 is insufficient.
If we combine the statements we are constrained by Statement 2 to using only numbers between 899 and 998.
3 x 899 = 2697. So we know the hundreds digit of R cannot be 8. Can it be 9?
Given that the Statements on GMAT data sufficiency problems are supposed to work together, we could assume that at least one case has to work and therefore 9 has to be the hundreds digit, but lets confirm that it works.
3 x 900 = 2700. So we just need R to be enough over 900 that multiplying R by 3 generates a number between 2800 and 2899. 3 x 950 = 2850. So R could be 950 and we have our hundreds digit for R.
Choose
C.
