Question of the Week - 27 (q is a three-digit number, in which ......)

Question

e-GMAT          Question of the Week #27

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

EgmatQuantExpert · Accepted Answer

Solution 
  Given:  
 • q is a three-digit number
 o Let’s say q = abc 
• The hundreds digit of q > the units digit of q
 o a > c 
• The units digit = the tens digit
 o c = b 
•  10^{99} – q  is divisible by 9

To find:  
 • The number of possible values of q

Approach and Working:   
 • We are given that  10^{99} – q  is divisible by 9
 o This implies, the sum of the digits in  10^{99} – q = 9k , where k is an integer.

• The sum of the digits of q = a + b + c
 o And, we can say that 1 ≤ a + b + c ≤ 27, since, a, b and c are single digit numbers .......... (1)  
 
 • Now, if we observe, the sum of the digits in  10^{99} = 1 , or we can consider it as 100
 o This implies, 100 – (a + b + c) = 9k .............. (2)

From (1) and (2), we can say that, (a + b + c) can be either 1 or 10 or 19
 • And we know that b = c. So, a + 2b = 1 or 10 or 19 and a must be > b

Case 1 : a + 2b = 1
 • Since, a > b, we get, a = 1 and b = 0
• Thus, in this case, 100 is the only possible value of q

Case 2 : a + 2b = 10
 • If b = 1, then a = 10 – 2 = 8, which is greater than b = 1. Thus, q = 811
• If b = 2, then a = 10 – 4 = 6, which is greater than b = 2. Thus, q = 622
• If b = 3, then a = 10 – 6 = 4, which is greater than b = 3. Thus, q = 433
• If b = 4, then a = 10 – 8 = 2, which is less than b = 4. Thus, this case is not valid

And if we go further, the value of ‘b’ increases and the value of ‘a’ decreases. So, all those cases are not valid.

So, the possible values of q in this case are {433, 622, 811}

Case 3 : a + 2b = 19
From this we can infer than a must be odd, since, odd + even = odd
 • If a = 9, then b =  \frac{(19 – 9)}{2} = 5 , which is less than a = 9. Thus, q = 955

• If a = 7, then b =  \frac{(19 – 7)}{2} = 6 , which is less than a = 7. Thus, q = 766

• If a = 5, then b =  \frac{(19 – 5)}{2} = 7 , which is greater than a = 5. Thus, this is not valid

And if we go further, the value of ‘a’ decreases and the value of ‘b’ increases. So, all those cases are not valid.

So, the possible values of q in this case are {766, 955}

Therefore, the total possible values of q = {100, 433, 622, 766, 811, 955}

Hence the correct answer is Option C.

Answer: C

https://e-gmat.com/blogs/wp-content/uploads/2018/07/FT-GC-FB-event-e1532524398949.jpg

Archit3110 · Answer

EgmatQuantExpert  : my solution took me >2 mins.. please provide official solution

given relation Q=ABC
and A>C and B=C

Q=100A+10B+C
or say 100A+11B

we need to check  10^{99} – q  values of q we get integer

so as to get an integer value the sum of digits of series Q should add up to 10 , which upon subtraction from 10^99 would give a number divisible by 9

so from the list of three digit no following would be value of Q

100, 433,622,811 , 955, 766

IMO C is correct...

Srikantchamarthi · Answer

Archit3110  do you mind explaining it again especially 766, 955

Archit3110 · Answer

reason for 766, 955 so as to get an integer value the sum of digits of series Q should add up to 10 , which upon subtraction from 10^99 would give a number divisible by 9

so in this case 766 ; 7+6+6=19= 1+9= 10 and 955; 9+5+5= 19=1+9=10

Hope this helps..

Srikantchamarthi

AdityaHongunti · Answer

If someone cannot come up with the above mentioned approaches , you can simply start writing down the numbers which wont take long..
but firat you to understand that the 10^99 is just there to intimidate you....all you have to consider is 1000 because we are subtracting a three digit integer ..if we were subtracting two digit integer then we would have consider 100 ..
why so?? seewhen we subtract a small number from such a huge number the initial digits are going to be the same 
for eg: 1000-100 = 900(initial number =9) 10000-100= 9900(initial num,bers 99) 100000-100= 99900(initial numbers =9)

so all we care about are the last three digits becasue those are the ones which are likely to be different from the rest of the initial numbers(in this case)...if we were subtracting a 4 digit number then we would only care for the last 4 digits of the bigger number becasue only those are likely to be different from other digits

for eg : 100000-100 = 99900 ... 100000-126= 99874 ... 100000-758 = 99242 (you can see that when you subtract 3 digit integer froma much larger number only the last three digits are likely to change)..you can do the same for subtraction of 4 digit number from a huge number

so the condition is : hundred digits( y) is greater than unit digit(x) ..and units and tens digit are same 
therefore = 100y +10x+ x = 100y +11x

now we have to subtract the 3 digit number from 1000(assumed as explained above becasue every digit beyond the thousands digit to the left is going to be 9)
so write dwon all the digits first ; 0,1,2,3,4,5,6,7,8,9
remember the condition that : hundred digit is greater than units : y>x
we start with 0 - we want a three digit integer
start with hundered digit as 1 = 100 (only one number becasue then unit digit would be greater or equal to 1)
with 2: 200, 211 
with 3 
:300,311,322
similary till hundred digit as 9 : 900,911,922,933,....988

one more way to refine your search is as soon as you get a number subtract it from 1000 : for eg "100" ..ill get a 9 ad 0 ,0 at the end ...so yeah divisible by 9
consider 311 : ill get a 6+8+9 = 23 : not divisible ...as you proceed youll get a rouh idea even befr=ore you subtract...

i did in this way and solved it in 2.39 seconds...

Fdambro294 · Answer

q = 100a + 10b + c

a > c ; a > b ; b = c

Rule I - you can Split Expressions and Subtract the Remainders

(10^99th / 9) Rem - (q / 9) Rem = Rem of 0

Any Power of 10 Divided by 9 will give you a Remainder = 1.

Since we are Subtracting q ---- we need a Number that will give us a Remainder of 1.

This will take the Form of: q = 9k + 1 ---- *where k = any Integer

Rule II - you can use the Divisibility Rule of 9 to find the Remainder. If the Digits Sum up to 9, the No. is divisible by 9. If the Digits Sum up to 10, the No. leaves a Remainder of 1 when divided by 9. If the Digits Sum up to 11, the No. leaves a Remainder of 2 when divided by 9. etc.

Thus, the Digits of q must SUM to = 9k + 1 ; so we get our Remainder of 1 to Subtract from (10^99th / 9) Rem

a + b + c = 9k + 1

Since b = c (Tens Digit = Units Digit)

a + 2b = 9k + 1

Given a > b, Start with a = 9 and use digits 0 thru 8 as = b to see whether you can find = 1 , 10 or 19 (All Multiples of 9 + 1)

Example Chart for 9:

9 + 2(8) = 25; NO
9 + 2(7) = 23; NO
at this point you should see a Pattern

9 + 2(5) = 19; YES ----- 955 is a Valid Value of q
.....
9 + 2(0) = 9; NO

Follow the Same path with Hundreds Digit = a = 8 ; 7 ; 6 ; 5 ; 4 ; 3 ; 2 ; 1

The Values that will work for q will be: 955; 811 ; 766 ; 622 ; 433 ; 100

Answer Choice C - 6 Values of q

Bambi2021 · Answer

10^99/9 will have a remainder of 1, thus q must also have a remainder of 1.

x = hundreds digit
y = tens digit = units digit

x + 2y = 9t + 1

From here I had to test numbers:

t = 0, then xyy = 100
t = 1, then xyy = 811, 622, 433
t = 2, then xyy = 955, 766

RamseyGooner · Answer

Can be solved using remainder theorem -
10^99 - q should be divisible by 9

10^99 | 9 -> remainder 1

so q | 9 -> should be 1 or 10 or 19 or 28 in order to be divisible by 9 -------1

we know q = htu and h> u = t

lets add up all possible u & t numbers -> 00, 11, 22, ..., 88 (99 not possible)
place the digits for h and try to matchback with eqn 1, we will get 6 combinations.

took me 2 mins

Hoe it helps!

danbrown9445 · Answer

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 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 = a + 2b 3.a>b 4.D is of the form 9k + 1 ......from (1) Largest value of q will be 988. So max D = 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 Reject because a Reject because a

versewriter · Answer

Option C i.e. total 6 those numbers.

As we know hundreds digit is greater than the units digit, and the units digit is equal to the tens digit. So If a number is divisible by 9 the sum of all digits of a number must also be divisible by 9. We know that 10^99 – q is divisible by 9. This means that -
10^99 not divisible by 9 as the sum of digits is divisible by 9, so if we put some value to q so that the resultant sum of digits will be divisible by 9. Then we can find the such total no of numbers that can be subtracted from 10^99 so that it can be divisible by 9, So -
eg- 10^99 - 1 = 99999999............99 (Total 99 zeros) will be divisible by 9
By using the above we can find the total such number.

Now such number must satisfy the condition of '' hundreds digit is greater than the units digit, and the units digit is equal to the tens digit'' . This means that the sum must be 10 of all such number so that if we subtract that number with the 10^99, the resultant will be divisible by 9. So such number will be total 06 i.e. 100, 433, 622, 811,955, 766

So answer is option C

bumpbot · Answer

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Question of the Week - 27 (q is a three-digit number, in which ......)

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Question of the Week - 27 (q is a three-digit number, in which ......)

Prep Toolkit

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My Rewards