Author 
Message 
TAGS:

Hide Tags

Director
Joined: 01 Apr 2008
Posts: 828
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
Updated on: 06 Nov 2012, 03:52
Question Stats:
60% (00:46) correct 40% (00:55) wrong based on 1228 sessions
HideShow timer Statistics
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ? A. 15 B. 16 C. 17 D. 18 E. 19
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by Economist on 24 Mar 2009, 23:51.
Last edited by Bunuel on 06 Nov 2012, 03:52, edited 2 times in total.
Renamed the topic and edited the question.



Manager
Joined: 09 Nov 2008
Posts: 148

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
25 Mar 2009, 00:16
C:17
I brute forced this one. All the multiples of 3 + 1 will have a remainder of 1: 4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49  16 numbers total But then I thought, no way it's this easy and thought about 1. 1/3 would also have a remainder of 1, making the answer 17.



Manager
Joined: 02 Mar 2009
Posts: 123

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
25 Mar 2009, 01:10
The range of numbers that leave a remainder of 1: 149
Therefore total number of integers: (491)/3 + 1 = 17
The last +1 is because the question is inclusive.
Hence C



Intern
Joined: 07 Feb 2009
Posts: 46

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
25 Mar 2009, 01:21
My ans is also C.17.
Explanation:
1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on. Hence we have an arithmetic progression: 1, 4, 7, 10,..... 49, which are in the form 3n+1. Now we have to find out number of terms. tn=a+(n1)d, where tn is the nth term of an AP, a is the first term and d is the common difference. so, 49 = 1+(n1)3 or, (n1)3 = 48 or, n1 = 16 or, n = 17



Math Expert
Joined: 02 Sep 2009
Posts: 46280

How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
14 Jun 2010, 01:11
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?A. 15 B. 16 C. 17 D. 18 E. 19 Algebraic way:Integer have a remainder of 1 when divided by 3 > \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ... \(n=3p+1\leq{50}\) > \(3p\leq{49}\) > \(p\leq{16\frac{1}{3}}\) > so \(p\), can take 17 values from 0 to 16, inclusive. Answer: C.
_________________
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



Manager
Joined: 05 Mar 2010
Posts: 182

Re: Help!!! [#permalink]
Show Tags
14 Jun 2010, 21:53
Does it mean that if zero is included in any similar problem like this then we should consider it, no matter what the divisor is??? Ex if question say any no. between 0 and 50, inclusive, divisible by 3 Answer will still be 17 Is it right
_________________
Success is my Destiny



Math Expert
Joined: 02 Sep 2009
Posts: 46280

Re: Help!!! [#permalink]
Show Tags
14 Jun 2010, 22:05



Manager
Joined: 05 Mar 2010
Posts: 182

Re: Help!!! [#permalink]
Show Tags
14 Jun 2010, 22:40
Good to know something new every time i login on this forum Thanks
_________________
Success is my Destiny



Intern
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 32
Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate  Thomson Reuters
WE 2: Analyst  TIAA CREF

Re: Help!!! [#permalink]
Show Tags
15 Jun 2010, 11:54
bibha wrote: How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3? A.14 B.15. C.16 D.17 E.18 ((Last  First)/ n) +1 > (491)/3 +1 = 17 We use 49 because that is the last that will produce a remainder of 1 when divided by 3 and 1/3 has a remainder of 1.



Manager
Joined: 15 Mar 2009
Posts: 144
Concentration: Finance, Entrepreneurship
Schools: UCLA (Anderson)  Class of 2014

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
22 Jul 2010, 21:13
My answer is C.
There are 48/3=16 numbers that are divisible by 3 (0 is excluded). If we add 1 to each and every of these number, we still have 16 numbers that have a remainder of 1 when divided by 3 (say 4, 7, 10,..., 49). But we have not counted 0 yet, 0+1 equals 1, 1 has a remainder of 1 when divided by 3.
Hence C.



Manager
Joined: 06 Apr 2010
Posts: 74

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
31 Aug 2010, 23:56
I think we can find this very quickly by using this: approximately 1/3 of the numbers from 050 are div by 3 so, (50/3) +1 = 16 + 1 = 17 approximately 1/2 of the numbers are div by 2 => (50/2 ) + 1 = 26 approximately 1/4 of the numbers are div by 4 => (50/4) + 1 =13 approximately 1/5 of the numbers are div by 5 => (50/5) + 1 =11 approximately 1/6 of the numbers are div by 6 => (50/6) + 1 = 9 I saw this on Bunuel's post somewhere and I thought to myself, how did he know? so I tried to calculate it and voila!, it is true...didn't try with range starting with nonzero numbers though don't forget to add one..before you are done!



Manager
Joined: 17 Mar 2010
Posts: 156

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
01 Sep 2010, 00:12
when 1 is devided by 3 remainder is 1, and when 3 is devided remainder 0. 16*3=48. so upto 48 there will be 16 number which will give remainder 1 after 48, there is only 1 number up to 50 which will give 1 remainder so the answer 17



Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 545

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
06 Jan 2012, 06:22



Manager
Joined: 29 Jul 2011
Posts: 98
Location: United States

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
06 Jan 2012, 20:57
First, lets look at the range. With 1 remainder, 449. So, number of elements = (493)/3 + 1 = 16 Now, lets not forget 1, since 1/3 > 1 as remainder (Good one!) So, 16 +1 = 17
_________________
I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!
DS  If negative answer only, still sufficient. No need to find exact solution. PS  Always look at the answers first CR  Read the question stem first, hunt for conclusion SC  Meaning first, Grammar second RC  Mentally connect paragraphs as you proceed. Short = 2min, Long = 34 min



Intern
Joined: 23 May 2012
Posts: 30

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
18 Oct 2012, 20:57
General Formula : n= px+q p=3 q=1 n=3x+1 Substituting Values n=0,4,7...49 . x = 0 to 16, inclusive Hence Total nos : 17



Intern
Joined: 22 Jan 2012
Posts: 20

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
04 Feb 2013, 23:48
Bunuel wrote: Algebraic way:
Integer have a remainder of 1 when divided by 3 > \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ...
\(n=3p+1\leq{50}\) > \(3p\leq{49}\) > \(p\leq{16\frac{1}{3}}\) > so \(p\), can take 17 values from 0 to 16, inclusive.
hi, i was wondering how we got 17 from the calculation that \(p\leq{16\frac{1}{3}}\) ? my approach was: 500+1 = 51 integers total 0/3 has r = 0 1/3 has r = 1 2/3 has r = 2 3/3 has r = 0 ... and so on thus there will be 1 value for every three that will have a remainder of 1 when divided by 3 (cyclicity of 3?) so 51/3 = 17 would this method work for similar questions? i was wondering how we treat the \(p\leq{16\frac{1}{3}}\) term when trying the algebraic method? i.e. how do we know to arrive at 17 ? thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 46280

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
05 Feb 2013, 04:00
essarr wrote: Bunuel wrote: Algebraic way:
Integer have a remainder of 1 when divided by 3 > \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ...
\(n=3p+1\leq{50}\) > \(3p\leq{49}\) > \(p\leq{16\frac{1}{3}}\) > so \(p\), can take 17 values from 0 to 16, inclusive.
hi, i was wondering how we got 17 from the calculation that \(p\leq{16\frac{1}{3}}\) ? i was wondering how we treat the \(p\leq{16\frac{1}{3}}\) term when trying the algebraic method? i.e. how do we know to arrive at 17 ? thanks! We have that \(n=3p+1\), where \(p\) is an integer \(\geq{0}\). So, n can be 1 (for p=0), 4 (for p=1), 7, 10, ..., and 49 (for p=16) > 17 values for p > 17 values for n. OR: \(p\leq{16\frac{1}{3}}\) implies that p can take integer values from 0 to 16, inclusive, thus it can take total of 17 values. Hope it's clear.
_________________
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



Intern
Joined: 18 Oct 2011
Posts: 3

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
06 Feb 2013, 03:24
The range of given numbers is 500+1=51 Any number when divided by 3 can give remainders from the list {0,1,2}. This patten will repeat for every 3 numbers. Hence divide the range by number of available remainders. i.e 51/3=17



Intern
Joined: 01 Apr 2013
Posts: 22

Re: PS: 0 to 50 inclusive, remainder [#permalink]
Show Tags
06 May 2013, 05:41
abhishekik wrote: My ans is also C.17.
Explanation:
1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on. Hence we have an arithmetic progression: 1, 4, 7, 10,..... 49, which are in the form 3n+1. Now we have to find out number of terms. tn=a+(n1)d, where tn is the nth term of an AP, a is the first term and d is the common difference. so, 49 = 1+(n1)3 or, (n1)3 = 48 or, n1 = 16 or, n = 17 Hello, I like this solution using the arithmetic progression but in the expression tn=a+(n1)d why n1 and not n ? As I know Un= U0 + n r where U0 is the first terme and r is the common difference. Thx in advance for the lighting !
_________________
The Kudo, please



Current Student
Joined: 22 Apr 2015
Posts: 49
Location: United States
GPA: 3.86

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Show Tags
01 Jul 2015, 16:06
I chose C
Since 3 has 16 multiples b/t 050 there must be 16 with a remainder of 1. But also \(1/3\) is included in this list. so 16+1=17=C




Re: How many integers from 0 to 50, inclusive, have a remainder
[#permalink]
01 Jul 2015, 16:06



Go to page
1 2
Next
[ 30 posts ]



