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Question of the Week  27 (q is a threedigit number, in which ......)
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14 Dec 2018, 02:52
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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?
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Question of the Week  27 (q is a threedigit number, in which ......)
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Updated on: 17 Dec 2018, 02:58
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... EgmatQuantExpert wrote: 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?
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Originally posted by Archit3110 on 14 Dec 2018, 05:29.
Last edited by Archit3110 on 17 Dec 2018, 02:58, edited 2 times in total.



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Re: Question of the Week  27 (q is a threedigit number, in which ......)
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14 Dec 2018, 07:05
811 622 433 955 766 100



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Re: Question of the Week  27 (q is a threedigit number, in which ......)
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15 Dec 2018, 19:38
Archit3110 do you mind explaining it again especially 766, 955



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Question of the Week  27 (q is a threedigit number, in which ......)
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15 Dec 2018, 23:49
Srikantchamarthi wrote: Archit3110 do you mind explaining it again especially 766, 955 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
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Re: Question of the Week  27 (q is a threedigit number, in which ......)
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17 Dec 2018, 01:58
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: 1000100 = 900(initial number =9) 10000100= 9900(initial num,bers 99) 100000100= 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 : 100000100 = 99900 ... 100000126= 99874 ... 100000758 = 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...



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Re: Question of the Week  27 (q is a threedigit number, in which ......)
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17 Dec 2018, 14:00



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Question of the Week  27 (q is a threedigit number, in which ......)
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19 Dec 2018, 06:44
Solution Given:• q is a threedigit number • The hundreds digit of q > the units digit of q • The units digit = the tens digit • \(10^{99} – q\) 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
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Question of the Week  27 (q is a threedigit number, in which ......) &nbs
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19 Dec 2018, 06:44






