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.

Let the q = 100x + 10y +z
where x >z and y = z
Other condition: \(10^{99}\) - q is divisible by 9
=> Remainder when 10^{99} divided by 9 : \(\frac{1^{99}}{9}\)= 1 (Remainder)
So, the \(\frac{q}{9}\) should also have 1 as a remainder.
then only the number is divisible by 9.

Thus, going through possible combinations, we get following numbers which when divided by 9 leaves a remainder of 1:
100,433,622,766,811,955

Thus q can have 6 possible values.

Please hit kudos if you like the solution.

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

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Please help:

If 10^99q needs to be divisible by 9 -> q must be divisible by 9.

Which means the digits of q must add up to a multiple of 9. The possible numbers with yxx format (y>x) are 522, 711, 855, 900 -> 4 numbers. What am I doing wrong here? TIA

No matter how I try to solve this, I have to look into all the possible ways the number XYY can be written, which is too time consuming.

Is this even GMAT worthy ?

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

Hi,

The shortest way to solve this exercise is to identify first the remainder for 10^99 when divided by 9, obviously it is 1.
Therefore, q will also have a remainder of 1 when divided by 9 because we want a number divisible by 9, so we subtract a remainder of 1 from a remainder of 1. Which means q must add up to 10 or to 19.

We already have the constraints of the 2 last digits that have to be the same and B>A :

BAA

when q= 100, remainder is 1
when AA=11 B=8 (A+A+B=1+1+8=10) therefore 811/9 gives a remainder of 1
when AA=22 B=6 (2+2+6=10)
when AA=33 B=4 (3+3+4=10)
Now when AA=44 B=2 (4+4+2=10) ==> however, we can't do that as B>A
when AA=55 B=9 (5+5+9=19) => 19/9 gives a remainder of 1
when AA=66 B=7 (6+6+7=19)

That's it, we have a total of 6 possible combinations
Therefore Answer C)