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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # How many integers from 0 to 50, inclusive, have a remainder  Question banks Downloads My Bookmarks Reviews Important topics
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TAGS: Founder  Joined: 18 Apr 2015
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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Expert's post 00:00

Question Stats: 59% (00:54) correct 40% (01:19) wrong based on 27 sessions
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
[Reveal] Spoiler: OA

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Eddie Van Halen - Beat it R.I.P.

Founder  Joined: 18 Apr 2015
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Kudos [?]: 3511 , given: 12566

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
Expert's post
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Eddie Van Halen - Beat it R.I.P.

Intern Joined: 17 Aug 2020
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Kudos [?]: 29 , given: 21

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
I did it in a tedious way:
The integers that have a remainder of 1 when divided by 3 are of the form 3k+1 where k is an integer. We want to find the highest k such that 3k+1<=50
Now 50=3*16+2, however 49=3*16+1. So the last k that satisfies the condition is k=16.
every number of the form 3k+1 with k from 0 to 16 satisfies the condition --> there are 17 numbers like this. Intern Joined: 15 Aug 2020
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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if a number when divided by 3 and has reminder of 1 then the number is written as 3K+1, so we can consider all the multiples of 3 which are less than 50(because if 50 is considered then 50+1 is 51, which is 51), so numbers will be 0, 3,6...48(which are multiple of 3 less than 50) and last multiple is 48, that is 3*16, so total count is 17(including 0). Hence Answer is C Moderator Joined: 20 Aug 2020
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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Carcass wrote:
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

Remind that the number of terms in an arithmetic sequence is $$\frac{l-f}{d} + 1$$, where $$l, f$$ and $$d$$ are its last term, it first term and its common difference, respectively.

The numbers between $$0$$ and $$50$$, inclusive with the remainder $$1$$, when divided by $$3$$ are $$1, 4, 7, ... , 49$$, which is an arithmetic sequence.

Thus, the number of terms is $$\frac{49-1}{3}+1 = \frac{48}{3}+1 = 16+1 = 17$$.

Therefore, the right answer is C. GRE Instructor Joined: 10 Apr 2015
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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Expert's post
Carcass wrote:
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

Although there are different (and clever) ways to answer this question, I think the fastest (and most accurate) approach is to simply list the values in your head as you count on your fingers (or use a tally sheet)

We get: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.  Intern Joined: 25 Aug 2020
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Kudos [?]: 6  , given: 0

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
1
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If you want to solve problems like these under 1 minute then always approach it by dividing the number by the given integer

how to know the pattern
3*1 + 1= 4 (rem = 1when div by 3)
3*2 + 1 = 7 (rem = 1when div by 3)
.....

50/3= 16.667
so we know 3*16 = 48 (hence 16 numbers are to be there which will give rem = 1) because 48 + 1 is < 50
finally add 1 because it'll give the same rem so Re: How many integers from 0 to 50, inclusive, have a remainder   [#permalink] 07 Oct 2020, 18:28
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