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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

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 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

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^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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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.

Re: The sum of all the digits of the positive integer q is equal [#permalink]

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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

Re: The sum of all the digits of the positive integer q is equal [#permalink]

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17 Aug 2012, 23:24

1

This post received KUDOS

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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\(10^n\) 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?

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 n-2 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.

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.

Re: The sum of all the digits of the positive integer q is equal [#permalink]

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18 Jul 2013, 11:59

2

This post was BOOKMARKED

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.(n-2) + (5+1) = 9.(n-2) + 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.

Re: The sum of all the digits of the positive integer q is equal [#permalink]

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19 Jul 2013, 15:36

100-49= 51 1000-49=951 10000-49=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

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 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 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

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 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.

We got that the sum of three-digit 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.
_________________

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 \(10^n-49\) 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 --> \(x=2\) --> \(9n=225\) --> \(n=25\).

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.

how come 9n-12 = x13 ---> 9n = x25 ??

Three-digit number x13 plus 12 is three-digit number x25. For example, 113 + 12 = 125.
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