EgmatQuantExpert wrote:
e-GMAT Question of the Week #27q 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?
The sum of digits of 10^3 and 10^99 is same,so we can consider 10^3 instead of 10^99 to make finding a pattern easier.
Let x = \(10^{3} – q\).
x is divisible by 9.So q can be 991,982,973 etc.
Notice that the sum of digits of q , D[q] is of the form 9k + 1 ...........(1)
Testing with constraints:
1.q is 3 digit number
2.q is of the form abb.So D[q] = a + 2b
3.a>b
4.D[q] is of the form 9k + 1 ......from (1)
Largest value of q will be 988.
So max D[q] = 25 (9+8+8)
There are only 3 numbers of the form 9k+1 from 1 to 25.They are 1,10,19.
So a + 2b = 1 or a + 2b = 10 or a + 2b = 19
Testing values where a is 9 to 1
So 9 + 2b = 1 or 9 + 2b = 10 or 9 + 2b = 19
Only 9 + 2b = 19 will proved us with an integral value of b.
So first number is a =9,b=5
955
When a = 8.
So 8 + 2b = 1 or 8 + 2b = 10 or 8 + 2b = 19
Only 8 + 2b = 10 will proved us with an integral value of b.
So 2nd number is a =8 ,b=1
We can see a pattern here.When a is odd then a + 2b must also be odd in order to yield a non-negative integer.Likewise for even
7 + 2b = 19
766
6 + 2b = 10
622
5 + 2b = 19
577 -> Reject because a<b
4 + 2b = 10
433
3 + 2b = 19
388 -> Reject because a<b
2 + 2b = 10
244 -> Reject because a<b
With 1,there is only 1 possible number,100
Set is {955,811,766,622,433,100}