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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

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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=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

2

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
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

\(10^n\) 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

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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.
_________________

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

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11 Jul 2015, 13:40

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?

Sum of all digits of \(q=9(n-2)+5+1\) (since \(q=10^n-49=99..951\) or (n-2) of 9's) --> \(9(n-2)+5+1=x13\) (x13 is the three-digit number) --> \(9n-12=x13\) --> \(9n=x25\) (last two digits 13 added 12 = 25) --> last digit of \(n=5\) (so that 9*5 yields last digit 5 of x25)

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

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21 Mar 2016, 18:44

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

the last 2 digits must be 51. so sum 6. now..since q is 10^n - 49, it means that the rest of digits are 9. we can build an equation: 6+9y=X13 13*9 = 117. 117+6 = 123. so no. 23*9 = 207. 207+6 = 213. so works. we need to have 23 of 9 and last two 51. we can write it down..i did it just to make sure 9999999999999999999999951 since we have 25 digits, and if we add 49, then there would be 25 of zeroes. which means that n=25.

I almost made the mistake to consider that q is difference of two squares. I eliminated at first B and D..but later while solving the question, I realized that B is the answer...

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

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26 Dec 2016, 03:45

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.

Got confused with this step "x139n−12=x13 --> 9n=x259n=x25 -->" but got to know it's logic in the post mentioned at the bottom.
_________________

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.

Got confused with this step "x139n−12=x13 --> 9n=x259n=x25 -->" but got to know it's logic in the post mentioned at the bottom.

\(9n-12=x13\), so 9n - 12 equals to three-digit number x13, thus 9n = x13 + 12 = x25.
_________________

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