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q is a threedigit number, in which the hundreds digit is greater than
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10 Jul 2019, 00:10
Question Stats:
34% (02:31) correct 66% (03:30) wrong based on 73 sessions
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q is a threedigit 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
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WE: Information Technology (Computer Software)

Re: q is a threedigit number, in which the hundreds digit is greater than
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10 Jul 2019, 00:37
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.



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Re: q is a threedigit number, in which the hundreds digit is greater than
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10 Jul 2019, 03:33
RajatVerma1392, I really didn't get how you've figured out the possible solutions.
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Re: q is a threedigit number, in which the hundreds digit is greater than
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10 Jul 2019, 23:14
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



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Re: q is a threedigit number, in which the hundreds digit is greater than
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11 Jul 2019, 08:01
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 xa, 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.



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Re: q is a threedigit number, in which the hundreds digit is greater than
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06 Aug 2019, 19:15
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 ?



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Joined: 21 Feb 2017
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Re: q is a threedigit number, in which the hundreds digit is greater than
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28 Apr 2020, 03:52
@experts Bunuel chetan is this question representative of the GMAT and is there a shorter way of doing this?




Re: q is a threedigit number, in which the hundreds digit is greater than
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28 Apr 2020, 03:52




