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:

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.

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+(n-1)d, where tn is the nth term of an AP, a is the first term and d is the common difference. so, 49 = 1+(n-1)3 or, (n-1)3 = 48 or, n-1 = 16 or, n = 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.

an-I know that 51/17=3, in order to have a reminder +1, i will take x=16 an=16*3+1=49

Could you please explain this in detail??

I have meant that 51=3*x, than x=17, since we have limits from 0 to 50 inclusive the last number in sequence is 49 which is =16*3+1, as 50=16*3+2.
_________________

well Bibha, I would like to expalin this in a logical manner rather than by mathematical expression. each no. which is 1 more than multiple of three would be in this series. so upto 50 there are 16 such no.s so correspondingly there are 16 no.s which are 1 more than a multiple of three and the last being 49. reply me if it is not clear.

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
_________________

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

0 is a multiple of every integer (except zero itself), so there are \(\frac{48-0}{3}+1=17\) numbers divisible by 3 in the range 0-50 inclusive (check this: totally-basic-94862.html?highlight=range).

But in original question 0 is not considered as one of the numbers: the lowest value of n is 1 (for p=0) and the highest value of n is 49 (for p=16), so total of 17 such numbers.
_________________

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.

I think we can find this very quickly by using this: approximately 1/3 of the numbers from 0-50 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 non-zero numbers though don't forget to add one..before you are done!

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

First, lets look at the range. With 1 remainder, 4-49.

So, number of elements = (49-3)/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 = 3-4 min

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

Show Tags

04 Feb 2013, 22:48

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:

50-0+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 ?

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.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...