Jun 29 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Jun 30 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Jul 01 08:00 AM PDT  09:00 AM PDT Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Jul 01 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 464
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
21 Jan 2012, 16:02
Question Stats:
53% (02:49) correct 47% (02:52) wrong based on 556 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730




Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Value of n
[#permalink]
Show Tags
21 Jan 2012, 16:18
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.
_________________




Senior Manager
Joined: 13 Aug 2012
Posts: 416
Concentration: Marketing, Finance
GPA: 3.23

Re: The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
21 Dec 2012, 07:29
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
_________________
Impossible is nothing to God.




Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 464
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

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^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;
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 55804

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



Manager
Joined: 12 Nov 2011
Posts: 65

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=n2 so you left with => 9*k+5+1=x13, n=k+2, find n, try numbers and solve it easy



Senior Manager
Joined: 24 Aug 2009
Posts: 459
Schools: Harvard, Columbia, Stern, Booth, LSB,

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



Intern
Joined: 20 Jun 2011
Posts: 44

Re: Value of n
[#permalink]
Show Tags
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.



Manager
Joined: 18 Oct 2011
Posts: 82
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01302013
GPA: 3.3

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 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.



Intern
Joined: 15 Jul 2013
Posts: 9

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.



Intern
Joined: 30 Apr 2010
Posts: 19

Re: The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
18 Jul 2013, 11:59
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.



Intern
Joined: 16 Jul 2013
Posts: 26

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



Intern
Joined: 10 Mar 2013
Posts: 11

Re: Value of n
[#permalink]
Show Tags
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.



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: Value of n
[#permalink]
Show Tags
12 Jun 2014, 04:56
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.
_________________



Intern
Joined: 09 Nov 2013
Posts: 1

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 \(10^n49\) 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 ??



Math Expert
Joined: 02 Sep 2009
Posts: 55804

The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
01 Dec 2014, 03:14
sukriti201 wrote: 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 ?? Threedigit number x13 plus 12 is threedigit number x25. For example, 113 + 12 = 125.
_________________



Intern
Joined: 22 Dec 2014
Posts: 34

The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
11 Jul 2015, 13:40
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? Sum of all digits of \(q=9(n2)+5+1\) (since \(q=10^n49=99..951\) or (n2) of 9's) > \(9(n2)+5+1=x13\) (x13 is the threedigit number) > \(9n12=x13\) > \(9n=x25\) (last two digits 13 added 12 = 25) > last digit of \(n=5\) (so that 9*5 yields last digit 5 of x25) > Answer: B.



Board of Directors
Joined: 17 Jul 2014
Posts: 2540
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
21 Mar 2016, 18:44
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 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...



Director
Joined: 26 Oct 2016
Posts: 626
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
26 Dec 2016, 03:45
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. 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.
_________________
Thanks & Regards, Anaira Mitch



Math Expert
Joined: 02 Sep 2009
Posts: 55804

Re: The sum of all the digits of the positive integer q is equal
[#permalink]
Show Tags
26 Dec 2016, 04:02
anairamitch1804 wrote: 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. 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. \(9n12=x13\), so 9n  12 equals to threedigit number x13, thus 9n = x13 + 12 = x25.
_________________




Re: The sum of all the digits of the positive integer q is equal
[#permalink]
26 Dec 2016, 04:02



Go to page
1 2
Next
[ 23 posts ]



