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# q is a three-digit number, in which the hundreds digit is greater than

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Intern
Joined: 08 Jul 2019
Posts: 5
Location: India
q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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10 Jul 2019, 01:10
3
00:00

Difficulty:

95% (hard)

Question Stats:

34% (02:48) correct 66% (03:26) wrong based on 50 sessions

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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

PS its 10^99 -q
Manager
Joined: 08 Jan 2018
Posts: 98
Location: India
GPA: 4
WE: Information Technology (Computer Software)
Re: q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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10 Jul 2019, 01:37
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.
Manager
Joined: 06 Feb 2019
Posts: 107
Re: q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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10 Jul 2019, 04:33
RajatVerma1392, I really didn't get how you've figured out the possible solutions.

Posted from my mobile device
Intern
Joined: 24 Oct 2017
Posts: 32
Location: India
GMAT 1: 710 Q48 V39
GPA: 3.39
Re: q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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11 Jul 2019, 00:14

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
Intern
Joined: 26 Oct 2010
Posts: 13
q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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11 Jul 2019, 09:01
2
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.
Manager
Joined: 27 Mar 2017
Posts: 63
Re: q is a three-digit number, in which the hundreds digit is greater than  [#permalink]

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06 Aug 2019, 20:15
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 ?
Re: q is a three-digit number, in which the hundreds digit is greater than   [#permalink] 06 Aug 2019, 20:15
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