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The sum of all the digits of the positive integer q is equal

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The sum of all the digits of the positive integer q is equal [#permalink] New post 21 Jan 2012, 16:02
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The sum of all the digits of the positive integer q is equal to the three-digit number x13. If q = 10^n – 49, what is the value of n?

(A) 24
(B) 25
(C) 26
(D) 27
(E) 28
[Reveal] Spoiler: OA

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Re: Value of n [#permalink] New post 21 Jan 2012, 16:18
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enigma123 wrote:
The sum of all the digits of the positive integer q is equal to the three-digit 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^n-49 will have n digits: n-2 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of 10^n-49 is equal to the three-digit number x13 --> 9(n-2)+5+1=x13 --> 9n-12=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] New post 21 Jan 2012, 17:03
Sorry Bunuel - Struggling to understand how you got to 10^n-49 will have n digits: n-2 9's and 51 in the end. For example: 10^4=10,000-51=9,949 --> 4 digits: 4-2=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] New post 21 Jan 2012, 17:14
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enigma123 wrote:
Sorry Bunuel - Struggling to understand how you got to 10^n-49 will have n digits: n-2 9's and 51 in the end. For example: 10^4=10,000-51=9,949 --> 4 digits: 4-2=two 9's and 49 in the end;


10^n-49 will have n digits: n-2 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

Or maybe this will help:
10^n has n+1 digits: 1 and n zeros;
10^n-49 has n digits, so one less digit than 10^n. In the end it'll have 51 (the same way 1,000-49=951) and the rest of the digits, so n-2 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] New post 22 Jan 2012, 08:42
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=n-2
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] New post 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^n-49 is equal to the three-digit number x13
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Re: The sum of all the digits of the positive integer q is equal [#permalink] New post 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] New post 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] New post 17 Aug 2012, 23:24
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 three-digit number which is x13.....} rather than {The sum of all the digits of the positive integer q is equal to the three-digit number x13.}
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28
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Re: Value of n [#permalink] New post 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^n-49 will have n digits: n-2 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;

We are told that the sum of all the digits of 10^n-49 is equal to the three-digit number x13 --> 9(n-2)+5+1=x13 --> 9n-12=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] New post 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 three-digit 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
Re: The sum of all the digits of the positive integer q is equal   [#permalink] 21 Dec 2012, 07:29
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