Author 
Message 
TAGS:

Hide Tags

VP
Status: mission completed!
Joined: 02 Jul 2009
Posts: 1297
GPA: 3.77

How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 00:50
Question Stats:
42% (02:57) correct 58% (02:28) wrong based on 241 sessions
HideShow timer Statistics
How many integers between 1 and 10^21 are such that the sum of their digits is 2? A. 190 B. 210 C. 211 D. 230 E. 231
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Audaces fortuna juvat!
GMAT Club Premium Membership  big benefits and savings




Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4493

How many integers between 1 and 10^21 are such that the sum
[#permalink]
Show Tags
04 Dec 2012, 14:47
How many integers between 1 and 10^21 are such that the sum of their digits is 2? (A) 190 (B) 210 (C) 211 (D) 230 (E) 231For a complete explanation, see: http://gmat.magoosh.com/questions/835When you submit your answer, the next page will have a complete video explanation. Each one of Magoosh's 800+ practice GMAT questions has its own video explanation, for accelerated learning.
_________________
Mike McGarry Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)




CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2605
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 03:49
E For the combinations of 2's and 0's we have one 2 and 20 0's total combinations =\(\frac{21!}{20!}\)= 21 > because 20 0's are identical. For the combinations of 1's and 0'2 we have two 1's and 19 0's Total combinations = \(\frac{21!}{19!*2!}\)= 210 Total = 210+21 = 231
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html




Retired Moderator
Joined: 02 Sep 2010
Posts: 769
Location: London

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 00:55
Pkit wrote: How many integers between 1 and \(10^21\) (10 in 21st power) are such that the sum of their digits is 2? a.190 b.210 c.211 d.230 e.231 Bunuel, please leave a chance for other people to solve this. thanks between 1 and 10^21 means we are talking about numbers with upto 21 digits. Case 1 Numbers of the form 2,20,200,etc Possible numbers = 21 (1 possibility for each number of digits) Case 2Numbers using the digits {1,1} Such a number will have a minimum of 2 digits and a maximum of 21 digits, with the first digit=1. Now, consider the kdigit number, with the first digit "1", out of the rest of the k1 digits, we have k1 ways to choose where to put the second 1. So k1, kdigit numbers possible. So total number of numbers = Summation(k1;k=2 to k=21)=1+2+...+20=20(21)/2=210 So total number of numbers = 210+21=231 Answer is (e)
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 13 Aug 2010
Posts: 166

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 03:15
so the second condition means there are 21 positions and two number, 1 and 1 so place in then. hence we can use the combination formula 21c11 and get 210.



Retired Moderator
Joined: 02 Sep 2010
Posts: 769
Location: London

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 03:25
Yea that's a good way to think about it But formula will be C(21,2) not C(21,11). C(21,2)=210 Posted from my mobile device
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 270
Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
16 Oct 2010, 05:33
Wow!! whenever I see these kinda questions I got stunned but guys like bunuel,shrouded1, and gurpreet are there to help . Nice method gurpreet and whenever a quant question is posted, shrouded1 is always there to explain it. Thank you! guys (I couldn't solve it but now i know how to solve it )
_________________
"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??



VP
Joined: 02 Jul 2012
Posts: 1184
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: the sum of the digits is 2
[#permalink]
Show Tags
04 Dec 2012, 21:39
Nice question..... 21C2 + 21 = 210 + 21 = 231
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Thanks To The Almighty  My GMAT Debrief
GMAT Reading Comprehension: 7 Most Common Passage Types



Manager
Joined: 26 Jul 2011
Posts: 94
Location: India
WE: Marketing (Manufacturing)

Re: How many integers between 1 and 10^21 are such that the sum
[#permalink]
Show Tags
05 Dec 2012, 02:55
Mike and Macfauz
are'nt there 22 places to choose for placing two 1s??



VP
Joined: 02 Jul 2012
Posts: 1184
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: How many integers between 1 and 10^21 are such that the sum
[#permalink]
Show Tags
05 Dec 2012, 03:13
ratinarace wrote: Mike and Macfauz
are'nt there 22 places to choose for placing two 1s?? \(10^{21}\) has 22 places. But we are asked for "between 1 & 10^21". The greatest integer less than 10^21 will have only 21 places.
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Thanks To The Almighty  My GMAT Debrief
GMAT Reading Comprehension: 7 Most Common Passage Types



Intern
Joined: 11 Sep 2012
Posts: 7

Re: How many integers between 1 and 10^21 are such that the sum
[#permalink]
Show Tags
28 Dec 2012, 10:10
Combinations of digits that sum up to 2 are either 1+1 or 2+0: Combinations 1+1:Double Digits11 Total: 1 Triple Digits101 110 Total: 2 Quadruple Digits1001 1010 1100 Total: 3 So now the pattern emerges that for every N digit number, there are N1 combinations summing up to two. The number of integers should be between 1 and 10^21, exclusive which means that the largest allowed integer does only have 21 digits, not 22 which means that the amount combinations with the largest amount of places is 20 (211). We can thus calculate the total amount of integers that satisfy the question by calculating the set of consecutive integers: 1+2+3...+20 which is 210. However, we did not account for the combinations of 2 and 0 that sum up to two, we have to add them first: Combinations 2+0:Single Digit2 Total: 1 Double Digits20 Total: 1 Triple Digits200 Total: 1 This pattern is obviously easier to comprehend and will account for an additional 21 combinations. Hence: 210+21 = 232. Sorry for my rusty English, I am not a native speaker .



Intern
Joined: 29 Jan 2015
Posts: 8
Concentration: Finance, Economics
GPA: 3.7

Hard counting problem? New way to solve??
[#permalink]
Show Tags
29 Jan 2015, 11:52
Hello all. This is my first question. Hopefully someone could clarify.
Q: How many integers between 1 and 10^21 are such that the sum of their digits is 2?
My answer: : Every level has 2 digits when there are two 1's in the number.
Level 1 : 11 = 1 Level 2 : 101 & 110 = 2 Level 3 : 1001 , 1010 & 1100 = 3 Level 4 : 10001, 10010, 10100 & 11000 = 4
So apparently every level you rise, 1 more integer with sum of digits 2 is added.
In this case,starting from the first "0" this number > 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer.
However, the OFFICIAL answer mentions this solution:
10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231.
Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer?
Any help is welcome, thanks in advance!!



Math Expert
Joined: 02 Sep 2009
Posts: 50004

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
29 Jan 2015, 11:54
JeroenReunis wrote: Hello all. This is my first question. Hopefully someone could clarify.
Q: How many integers between 1 and 10^21 are such that the sum of their digits is 2?
My answer: : Every level has 2 digits when there are two 1's in the number.
Level 1 : 11 = 1 Level 2 : 101 & 110 = 2 Level 3 : 1001 , 1010 & 1100 = 3 Level 4 : 10001, 10010, 10100 & 11000 = 4
So apparently every level you rise, 1 more integer with sum of digits 2 is added.
In this case,starting from the first "0" this number > 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer.
However, the OFFICIAL answer mentions this solution:
10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231.
Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer?
Any help is welcome, thanks in advance!! Merging similar topics. Please refer to the solutions above.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Aug 2009
Posts: 6961

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
29 Jan 2015, 19:56
[quote="JeroenReunis"]Hello all. This is my first question. Hopefully someone could clarify. Q: How many integers between 1 and 10^21 are such that the sum of their digits is 2? My answer: : Every level has 2 digits when there are two 1's in the number. Level 1 : 11 = 1 Level 2 : 101 & 110 = 2 Level 3 : 1001 , 1010 & 1100 = 3 Level 4 : 10001, 10010, 10100 & 11000 = 4 So apparently every level you rise, 1 more integer with sum of digits 2 is added. In this case,starting from the first "0" this number > 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer. However, the OFFICIAL answer mentions this solution: 10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231. Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer? Any help is welcome, thanks in advance!![/quote hi Jeroen ...your approach is ok but u r going wrong on two accounts although u r getting the correct ans 1) as far as your counting is considered for two 1's... the last part where u have considered 21 at level 21 is wrong.. in ur way there should be only 20 levels.. the reason is 10^21 is the lowest 21 level digit number..that is 10000....000. and one lower than it is 999..20 times so it cannot have 1000..20 times...01 2) so the ans is 1+2+...+20, which is 210.. now what about rest 21.. this is the second mistake.. what about digit 2 with all rest digits as 0 for eg 2,20,200.. 2*10^0,2*10^1,2*10^2,.. till 2*10^20=21 such numbers combine the two 210+21=231 s
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
29 Jan 2015, 23:12
JeroenReunis wrote: Hello all. This is my first question. Hopefully someone could clarify.
Q: How many integers between 1 and 10^21 are such that the sum of their digits is 2?
My answer: : Every level has 2 digits when there are two 1's in the number.
Level 1 : 11 = 1 Level 2 : 101 & 110 = 2 Level 3 : 1001 , 1010 & 1100 = 3 Level 4 : 10001, 10010, 10100 & 11000 = 4
So apparently every level you rise, 1 more integer with sum of digits 2 is added.
In this case,starting from the first "0" this number > 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer. The answer is exactly correct but you got lucky! You will go till Level 20, not level 21 because level 20 will have all 21 digit numbers. Level 21 has 22 digit numbers but 22 digit numbers will be greater than 10^21 because 10^21 is the smallest 22 digit number. Then how come you get the correct answer? Because you have to add 21 to the sum for these 21 numbers: 2 20 200 2000 20000 200000 .... There will be 21 such numbers. The sum of digits here too is 2. JeroenReunis wrote: However, the OFFICIAL answer mentions this solution:
10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231.
Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer?
Any help is welcome, thanks in advance!! THink about this: Is 020 an integer? Yes. It is 20. (0 on the left of an integer has no value.) If we think this way, we don't have to consider the numbers with different number of digits separately. We can say that all number have 21 digits such that 0 can be placed in the beginning too. 00000...0000000002 00000...0000000020 00000...0000000200 00000...0000002000 ... The same numbers as given above can be written like this. Makes sense?
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Senior Manager
Joined: 01 Nov 2013
Posts: 301
WE: General Management (Energy and Utilities)

How many integers between 1 and 10^21
[#permalink]
Show Tags
Updated on: 30 Jan 2015, 00:40
I think the options to the above questions fall short of the correct answer Here is my solution A 210^0 series1 (0+1) B 11,2010^1 series2 (1+1) C101,110,20010^2 series3 (2+1) D1001,1010,1100,200010^3 series4 (3+1) E10001,10010,10100,11000,2000010^4 series5 (4+1) and so on......................................................10^20 series..............................................21 (20+1) so essentially we have to find the sum of the series 1+2+3+................................+.20+21 = 231 Experts ..can u validate the above
_________________
Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.
I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.Mohammad Ali
Originally posted by samusa on 30 Jan 2015, 00:20.
Last edited by samusa on 30 Jan 2015, 00:40, edited 1 time in total.



Math Expert
Joined: 02 Aug 2009
Posts: 6961

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
30 Jan 2015, 00:30
samichange wrote: I think the options to the above questions fall short of the correct answer
Here is my solution
A 210^0 series1 (0+1) B 11,2010^1 series2 (1+1) C101,110,20010^2 series3 (2+1) D1001,1010,1100,200010^3 series4 (3+1) E10001,10010,10100,11000,2000010^4 series5 (4+1)
and so on......................................................10^21 series..............................................22 (21+1)
so essentially we have to find the sum of the series
1+2+3+................................+.21+22 = 253
Experts ..can u validate the above hi sami, u have done everything perfect except one point... 22 which u have added in10^21 is not to be added.. reason 10^21 is the only number and the lowest one in that series.. so u have to add 0 for that series.. ur ans will become 25322=231..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Senior Manager
Joined: 01 Nov 2013
Posts: 301
WE: General Management (Energy and Utilities)

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
30 Jan 2015, 00:38
chetan2u wrote: samichange wrote: I think the options to the above questions fall short of the correct answer
Here is my solution
A 210^0 series1 (0+1) B 11,2010^1 series2 (1+1) C101,110,20010^2 series3 (2+1) D1001,1010,1100,200010^3 series4 (3+1) E10001,10010,10100,11000,2000010^4 series5 (4+1)
and so on......................................................10^21 series..............................................22 (21+1)
so essentially we have to find the sum of the series
1+2+3+................................+.21+22 = 253
Experts ..can u validate the above hi sami, u have done everything perfect except one point... 22 which u have added in10^21 is not to be added.. reason 10^21 is the only number and the lowest one in that series.. so u have to add 0 for that series.. ur ans will become 25322=231.. Thanks!! i just realised the mistake ....
_________________
Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.
I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.Mohammad Ali



NonHuman User
Joined: 09 Sep 2013
Posts: 8467

Re: How many integers between 1 and 10^21
[#permalink]
Show Tags
28 Jul 2018, 11:47
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: How many integers between 1 and 10^21 &nbs
[#permalink]
28 Jul 2018, 11:47






