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The sum of all the digits of the positive integer q is equal [#permalink]
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21 Jan 2012, 16:02
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The sum of all the digits of the positive integer q is equal to the threedigit number x13. If q = 10^n – 49, what is the value of n? (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
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Re: Value of n [#permalink]
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enigma123 wrote: The sum of all the digits of the positive integer q is equal to the threedigit number x13. If q = 10^n – 49, what is the value of n? (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
Any idea how to approach this problem? \(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^n49\) will have \(n\) digits: \(n2\) 9's and 51 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^n49\) is equal to the threedigit number \(x13\) > \(9(n2)+5+1=x13\) > \(9n12=x13\) > \(9n=x25\) > \(x25\) is divisible by 9 > the sum of its digits must be divisible by 9 > \(x=2\) > \(9n=225\) > \(n=25\). Answer: B. Hope it's clear.
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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21 Jan 2012, 17:03
Sorry Bunuel  Struggling to understand how you got to 10^n49 will have n digits: n2 9's and 51 in the end. For example: 10^4=10,00051=9,949 > 4 digits: 42=two 9's and 49 in the end;
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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21 Jan 2012, 17:14
enigma123 wrote: Sorry Bunuel  Struggling to understand how you got to 10^n49 will have n digits: n2 9's and 51 in the end. For example: 10^4=10,00051=9,949 > 4 digits: 42=two 9's and 49 in the end; \(10^n49\) will have \(n\) digits: \(n2\) 9's and 51 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; Or maybe this will help: 10^n has n+1 digits: 1 and n zeros; 10^n49 has n digits, so one less digit than 10^n. In the end it'll have 51 (the same way 1,00049=9 51) and the rest of the digits, so n2 digits, will be 9's. Hope it's clear.
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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22 Jan 2012, 08:42
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My way after pluging n=2, n=3, n=4 you see that digits go according to the pattern "9(k times)51" where k=n2 so you left with => 9*k+5+1=x13, n=k+2, find n, try numbers and solve it easy



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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24 Jan 2012, 18:20
Bunuel  again struggling. Can you please explain how did you get to this ? We are told that the sum of all the digits of 10^n49 is equal to the threedigit number x13
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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24 Jan 2012, 18:20
I mean the solution after the line above.
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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24 Jan 2012, 18:22
Got it thanks. Sorry its bit late in the night at my end.
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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17 Aug 2012, 23:24
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Again the best solution by Bunuel. I think the question should be written as clear as possible to avoid any confusion. The stem of the question should be {The sum of all the digits of the positive integer q is equal to the threedigit number which is x13.....} rather than {The sum of all the digits of the positive integer q is equal to the threedigit number x13.} (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
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Re: Value of n [#permalink]
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14 Sep 2012, 07:56
Bunuel wrote: \(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros;
\(10^n49\) will have \(n\) digits: \(n2\) 9's and 51 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end;
We are told that the sum of all the digits of \(10^n49\) is equal to the threedigit number \(x13\) > \(9(n2)+5+1=x13\) > \(9n12=x13\) > \(9n=x25\) > \(x25\) is divisible by 9 > the sum of its digits must be divisible by 9 > \(x=2\) > \(9n=225\) > \(n=25\).
Answer: B.
Hope it's clear.
I'm just wondering about this approach: I tried to find a pattern. n= 2 > sum of digits = 6 n= 3 > sum of digits = 15 n= 4 > sum of digits = 24 n= 5 > sum of digits = 33 (an increase of 9 for every n) So for every odd n we get an odd units digit in the sum. And since x13 ends with an odd (3) number, I eliminated ans. choices A, C and E. As for B and D  I was guessing/hoping that the units digit would stay on 3 for multiples of 5. So I picked ans. choice B). But after thinking about it. I think I got lucky. So I'm wondering if there is something/similar that relates to this approach that yields a correct answer every time? Thanks.



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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21 Dec 2012, 07:29
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enigma123 wrote: The sum of all the digits of the positive integer q is equal to the threedigit number x13. If q = 10^n – 49, what is the value of n?
(A) 24 (B) 25 (C) 26 (D) 27 (E) 28 10^2  49 = 51 10^3  49 = 951 10^4  49 = 9951 The sum of digits 5 and 1 is 5 + 1 = 6. In order to get x13, we need to add a units digit 7 to 6. (A) 10^24 gives 22 9s = 9*22 = units digit 8 (B) 10^25 gives 23 9s = 9*23 = units digit 7 Answer: B
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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17 Jul 2013, 00:33



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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17 Jul 2013, 09:11
Let's say n = 2. This would leave q= 51. When n=3 this would leave q= 951. As n increases after this point a 9 will be added to the number. So for n=4 we would have 2 9's in the number. Following this pattern whatever n happens to be we will have n2 number of 9's left in our number q.
Since the 3 digit number we want is x13, the units digit of the addition of all the numbers in q must be a multiple of 9 + 6 (5 +1). If we look at the options given we can eliminate A as we would be left with (22x9) + 6 which would not yeild a units digit of 3. If we move to option B we can see that (23X9) + 6 does yeild a 3 in the units digit. If we test out the remaining options in this fashion we realize that only answer B gives us the desired result.



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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17 Jul 2013, 14:18
I'll give this a try......
x13 = sum of all the digits in our number (q), and q = 10^n – 49. So 10^any number will always end in a bunch of 0's, thus 10^n  49 will always be a bunch on 9's then end in 51, for example 9999999999999999951. So we know the last two digits are 5 and 1, and then every other digit is a 9. if we take x13 and subtract 5 and 1 (6) from it, then we get x07. Now we need to know how many 9's go into x07. If we add up all digits of x07, they need to equal 9 in order for 9 to be a factor of it, so x07 must really be 207. 207 = 9*23, so there are 23 9's in our number followed by a 5 and a 1, giving us a 25 digit number. 10^25 = a 26 digit number, subtract 49 yields a 25 digit number. Thus n=25.



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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18 Jul 2013, 11:59
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I also solved it by finding a pattern:
For n = 3: q = 10^3  49 = 951 For n = 4: q = 10^4  49 = 9951 For n = 5: q = 10^5  49 = 99951 etc...
We can see that the number of 9 digits is 2 less than n. We can now see that our number q equals 99999.....951. We can now test with the given values for n:
(A) n = 24: 9.(n2) + (5+1) = 9.(n2) + 6 = 9(22) + 6 = 204 (B) n = 25: (9)(23) + 6 = 213, our answer is (B)
I think this can be solved in under 2 minutes with this method if one can find the pattern quickly.



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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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19 Jul 2013, 15:36
10049= 51 100049=951 1000049=9951 ... sum the digits 5+1=6 9+6=15 2(9)+6=25 ... we want x(9)+6 to equal x13 the sum is obviously greater than 100, this means x>10 6 only adds to 13 when added with 7 and only 3*9=27 gives us the seven that we need we need the choices to be in multiples of 3, x=3,13, 23, 33, 43... gives you the ending 7 x=3, 27+6=33 x=13, 13*9+6=123 x=23, 23*9+6= 213
n=2+x, this is because we didn't see a 9 appear in 10^1, 10^2. thus...23+2=25



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Re: Value of n [#permalink]
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11 Jun 2014, 20:15
How do you get X = 2? My approach was after finding out 51, I started plugging in the numbers from the answers and all I was looking for is a number that sums up to a units digit of 3. Bunuel wrote: enigma123 wrote: The sum of all the digits of the positive integer q is equal to the threedigit number x13. If q = 10^n – 49, what is the value of n? (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
Any idea how to approach this problem? \(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^n49\) will have \(n\) digits: \(n2\) 9's and 51 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^n49\) is equal to the threedigit number \(x13\) > \(9(n2)+5+1=x13\) > \(9n12=x13\) > \(9n=x25\) > \(x25\) is divisible by 9 > the sum of its digits must be divisible by 9 > \(x=2\) > \(9n=225\) > \(n=25\). Answer: B. Hope it's clear.



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Re: Value of n [#permalink]
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12 Jun 2014, 04:56
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farhanabad wrote: How do you get X = 2? My approach was after finding out 51, I started plugging in the numbers from the answers and all I was looking for is a number that sums up to a units digit of 3. Bunuel wrote: enigma123 wrote: The sum of all the digits of the positive integer q is equal to the threedigit number x13. If q = 10^n – 49, what is the value of n? (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
Any idea how to approach this problem? \(10^n\) has \(n+1\) digits: 1 and \(n\) zeros. For example: 10^2=100 > 3 digits: 1 and 2 zeros; \(10^n49\) will have \(n\) digits: \(n2\) 9's and 51 in the and. For example: 10^449=10,00049=9,951 > 4 digits: 42=two 9's and 51 in the end; We are told that the sum of all the digits of \(10^n49\) is equal to the threedigit number \(x13\) > \(9(n2)+5+1=x13\) > \(9n12=x13\) > \(9n=x25\) > \(x25\) is divisible by 9 > the sum of its digits must be divisible by 9 > \(x=2\) > \(9n=225\) > \(n=25\). Answer: B. Hope it's clear. We got that the sum of threedigit number x25 is divisible by 9. A number is divisible by 9 if the sum of its digit is divisible by 9. So, for x25 to be divisible by 9 x must be 2: 2+2+5=9, which is divisible by 9.
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Re: The sum of all the digits of the positive integer q is equal [#permalink]
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30 Nov 2014, 09:25
Bunuel wrote:
We are told that the sum of all the digits of \(10^n49\) is equal to the threedigit number \(x13\) > \(9(n2)+5+1=x13\) > \(9n12=x13\) > \(9n=x25\) > \(x25\) is divisible by 9 > the sum of its digits must be divisible by 9 > \(x=2\) > \(9n=225\) > \(n=25\).
Answer: B.
Hope it's clear. how come 9n12 = x13 > 9n = x25 ??



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