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# How many integers from 0 to 50, inclusive, have a remainder

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Kudos [?]: 3525 [1] , given: 12608

How many integers from 0 to 50, inclusive, have a remainder [#permalink]  19 Aug 2020, 12:01
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62% (00:55) correct 37% (01:19) wrong based on 29 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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Founder
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Kudos [?]: 3525 [0], given: 12608

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  19 Aug 2020, 12:01
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Intern
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Kudos [?]: 29 [0], given: 21

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  20 Aug 2020, 03:53
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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Kudos [?]: 7 [1] , given: 0

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  20 Aug 2020, 05:33
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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
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Kudos [?]: 45 [1] , given: 0

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  20 Aug 2020, 09:46
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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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Kudos [?]: 4644 [1] , given: 70

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  03 Oct 2020, 06:15
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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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Intern
Joined: 25 Aug 2020
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Kudos [?]: 6 [1] , given: 0

Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]  07 Oct 2020, 18:28
1
KUDOS
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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