q is a three-digit number, in which the hundreds digit is greater than the units digit, and the units digit is equal to the tens digit. If \(10^{99} - q\) is divisible by 9, then what is the number of possible values of q?

A. 4
B. 5
C. 6
D. 9
E. 10

RajatVerma1392, I really didn't get how you've figured out the possible solutions.

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the question says that if (10^99 - Q) is divisible by 9, what are the possible values of Q.

So, we need to figure out the values of Q such that (10^99 - Q) is divisible by 9.

The above equation can be written as (10^99)/9 - Q/9. In other words, the reminder for equation must be 0.

Now, for the 1st term, we can use the reminder theorem. It states that if f(x) is divided by x-a, the reminder is f(a).

So, 10^99/9 can be written as 10^99 / (10 - 1). hence the reminder is f(1) = (1)^99 = 1. So, the reminder of the first term is 1.

Now, the reminder of the 2nd term (Q/9) should also be 1 so that the reminder of the entire equation is 0.

for Q we totally have 45 options: x00 to x88. among that, we need to find out in how many options the 3 numbers add to 10 or 19 or 28.

for x00: only 100
for x11: only 811
for x22: only 622
for x33: only 433
for x44: none
for x55: only 955
for x66: only 766
for x77: none
for x88: none

for totally 6 values.

Hope it is clear.

Bunuel, EgmatQuantExpert Please let us know if there is a shorter procedure. This is way too time consuming for the GMAT.

@experts Bunuel chetan is this question representative of the GMAT and is there a shorter way of doing this?