Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

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;
_________________

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]

Show Tags

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]

Show Tags

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
_________________

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]

Show Tags

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]

Show Tags

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]

Show Tags

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]

Show Tags

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]

Show Tags

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...