VJesus12 wrote: ↑Fri Jul 09, 2021 11:39 am
\(q\) 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
Answer:
C
Source: e-GMAT
Good question.
Step 1 is to realize that 10^99 is just to intimidate. In reality, any power of 10 when divided by 9 gives 1 as remainder.
Immediately, you would know that the remainder of q divided by 9 MUST be 1 so that 10^99 - q is divisible by 9 (i.e. remainder is 0).
So now, the initial part of the question. q is of the form ABB & A>B. Using 9s divisibility, we know that the sum of the digits of Q must be 1 more than multiples of 9
Possibilities: 1, 10, 19 (cant be more than that since 3*0 = 27)
A + 2B = 1
B=0 is only possibility. 100
A + 2B = 10
B = (10-A)/2 => A is even. take A=2, B=4 no. A=4, B=3 yes. A=6, B=2 yes. A=8, B=1, yes
433, 622, 811
A+2B = 19
B = (19-A)/2==> A is odd. A=1, B=9 no -> you can quickly see that A must be more than 5 since A>B. A=7, B=6, yes. A=9, B=5
766, 955
Total possibilities: 6.
You can also do the following way, but this will be more time consuming:
Possibilities:
A00 -> 100 (rest all are less than 9)
A11 -> 211 (quick sum 4, 311 would be 5 ....) only 811 possible
A22 -> 622
A33 -> 433 (derived by simply adding 1s to 2 and reducing 2 from 6)
A44 -> None since A>B
A55 -> and so on.
