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

Question

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

Show Answer

Official Answer

C

A. 4
B. 5
C. 6
D. 9
E. 10

sanjitscorps18 · Answer

10^99−q is divisible by 9

=> (10^99 - 1) - (q-1) is divisible by 9

Since (10^99 - 1) is divisible by 9, q-1 should be divisible by 9. Now q is of the form ABB where A > B

Hence possible values could be
q 433 , 955 , 766 , 622 , 800 , 100
q - 1 432 , 954 , 765 , 621 , 810 , 99

So, IMHO the answer should be Option C.

Edit - Missed 2 options earlier. Revised the answer.

100mitra · Answer

Correct Option C.

Number of possibility of q when (10^99 - q) divisible by 9.

Condition 1.   100 place > unit place and unit place is equal to 10th place.
A = 100 place digit 
B = unit and 10th place
Possible number structure must be  ABB

Condition 2 . 
10^99 / 9 = reminder 1
Q / 9 = reminder 1
 A + 2B = 9A + 1

A = 0, then ABB = 100
A = 1, then ABB = 811, 622, 433
A = 2, then ABB = 955, 766

Option C

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SheelzG · Answer

How you got the qeqn "A+2B=9A+1"? Can you explain? I am stuck with it..

100mitra · Answer

Sure

Happy to help. It's a fusion of Reminder Theron. 1. Dividend = Quotient* Divisor + reminder. D = dq + r Forget about anything divide by 10^power , simply consider if, 10 divided by 9, reminder will be 1. So we can write the equation. Where any 1 is quotient, can be a integer like A' 10 = 9*1 + 1 Same can be written as 10 = 9A' + 1 -----(1) reminder is 1 Coming back to original question. A is 100th Digit and B is tenth and C Unit digit. q = 100A + 10B + B a > C ; A > B ; B = C A + B + C = A + 2B. (As B = C) ----(2) ( 10^99 - q) is divisible by 9 ----(3) Since we are substrating q, it must be in such a way, reminder must be 1 to satisfy the condition (1) So the equation can be again written as \frac{10^99 }{ 9} - \frac{q}{9} = 0 To full fill the condition \frac{q}{9} and get reminder 1, q must be either 10, 19, 28, 37.... .and so on, but our condition gets satisfied till 19, no need to push beyond it. So q becomes 9*(multiple of 9) + reminder 1 Which I have simply stated as below, gives equation A + 2B = 9A' + 1 here A remains single digit and b has to be same 2 digit (as B = C) From here I had to test numbers: 1. A = 0, then ABB = 100 (100/9 = reminder 1) A = 1, then 2. ABB = 811 = (8+1+1)/9 = 10/9 reminder 1 3. ABB = 622 = (6+2+2)/9 = 10/9 reminder 1 4. ABB = 433 = (4+3+3)/9 = 10/9 reminder 1 A = 2, then 5. ABB = 955 = (9+5+5)/9 = 19/9 reminder 1 6. ABB = 766 = (7+6+6)/9 = 19/9 reminder 1 So totally we have 6 condition satisfied . Hope it's makes sense & helps you to clear your doubt

kingbucky · Answer

(10^99 - q) leaves reminder 1, when divided by 9, i.e. 
(10^99 - q) mod 9 =0
=> 10^99 mod 9 - q mod 9 =0
=> 1 mod 9 - q mod 9 =0
=> q=1 mod 9 i.e. q when divided by 9, leaves a reminder 1.

Let the number be of the form "mnn", where m>n

now,
q=1 mod 9
=> m+n+n =1 mod 9 (since the reminder when a number is divided by 9 is same as the reminder when its sum of digits is divided by 9)
=> m+2n =1 mod 9 => m+2n = 9k+1, where k is an integer

Now, since m,n >0
We have m+ 2n =1; m+2n =10; m+2n =19; or m+2n = 28 ...
Case (1) 
m+2n =1 :: yields (1,0) ie 100

Case (2) 
m+2n =10 ::
m > n
8     1  -> 811
6     2  -> 622
4     3  -> 433
2     4 -> m > n failed

Case (3) 
m+2n =19 ::
m > n
9     5  -> 955
7     6  -> 766
5     7  :: m>n failed
3     8  :: m>n failed
1     9  :: m>n failed

Case (4) 
m+2n =28 ::
here, Even setting m=n=9, only sums up to 27. Both cannot be single digit numbers.

Hence, from Case (1),(2) & (3)
We have
100
811
622
433
955
766

Aashimabhatia · Answer

10^99-Q is divisible by 9
We know 10^99 when divided by 9 leaves remainder 1, hence Q when divided by 9 will also leave remainder 1. 
This means, Q=9n+1 
Also, Q can be expressed in the form of ABB, where A>B 
For a number to be divisible by 9, its sum of digit should also be divisible by 9. In this case, Q is 1 more than multiple of 9, hence, sum of digits will also be 1 more than a multiple of 9.
So, A+B+B=9x+1
 A+2B=9x+1 
Now checking which values of Q satisfy this condition:
1. x=0, A+2B=1
only 100 satisfies this condition
2. x=1, A+2B=10
As A>B, we will put A=4,6,8 (only even integers as putting odd integers as A will render the value of B to be a fraction which is not possible given it should be a digit from 0 to 9)
433 
622
811
3. x=2, A+2B=19
766
955
4. x=3, A+2B=28
Not Possible as A & B are digits from 1 to 9, hence maximum value of ABB can be 9+8+8=25
No need to check further cases
So our required answer is 6 values (100, 433, 622, 811, 766, 955)

Ilanchezhiyan · Answer

Bunuel  Sir, is there an easier approach to arrive at the answer? What's your take on this?

eb590 · Answer

Hi can someone clarify this for me?

If A = 0 then 0+2B = 1 (mod 9)
2B = 2x5 = 10 = 1 (mod 9)

so B = 5

ABB = 055 why is ABB = 100 when A = 0

I understand 055 is not in the format ABB where A>B but I don't get how we come up with 100

Thank you for the clarification

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

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