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

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Intern
Joined: 30 Dec 2004
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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22 Mar 2007, 12:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

15
16
17
18
19

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Director
Joined: 12 Jun 2006
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22 Mar 2007, 12:50
0/3 = 0 R 0

1/3 = 0 R 1

2/3 = 0 R 2

3/3 = 1 R 0

4/3 = 1 R 1

5/3 = 1 R 2

6/3 = 2 R 0

By now you can see a pattern developing. 1 out of 3 remainders are 1. inclusive = (50 - 0) + 1
51/3 = 17

Does anyone have a quicker way?

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Intern
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22 Mar 2007, 13:47
this is an arithmetic sequence, a(n) = 3*n +1

where n can have values from 0 to 16
that is, 17 values.

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Director
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06 Apr 2007, 02:49
arithmetic sequence

1,4,7,......49

nth term = a+(n-1)d = 49
a=1
d=3
1+(n-1)*3 = 49
n = 17

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Senior Manager
Joined: 20 Feb 2007
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06 Apr 2007, 03:40
There are total 51 integers (0 to 50, inclusive). When we divide all of 51 integers by 3, we will get 17 integers with remainders 0, 17 integers with remainders 1 and 17 integers with remainders 2.

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VP
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06 Apr 2007, 05:10
a=1

last number =49 = 1 + (n-1)*3 [the last term in a AP is a+(n-1)*d]

n=17

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Manager
Joined: 20 Mar 2007
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Kudos [?]: 16 [0], given: 0

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09 Apr 2007, 09:15
ggarr wrote:
0/3 = 0 R 0

1/3 = 0 R 1

By now you can see a pattern developing. 1 out of 3 remainders are 1. inclusive = (50 - 0) + 1
51/3 = 17

Does anyone have a quicker way?

I do not understand the part 1/3 = O R 1.

I started this as an AP with first term = 4 last being 49 to get a total of 16 terms which when divided by 3 gives a remainder of 1.

How do you include '1' in this set to yeild 17?Are you considering 0 as a multiple of 3 to divide 1 by 3 to get the remainder as 1?

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Intern
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09 Apr 2007, 11:35
i get 16. you cannot include the numbers less than 3. that will not yeild a remainder of 1.

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Manager
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10 Apr 2007, 12:02
of course u can. 1 divided by 3 also has a remainder of 1.

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Director
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11 Apr 2007, 05:38
Yep, I see the easiest way to think of it is by looking at a sequence of numbers that have a remainder of 1 when divided by 3 and are less than 51.

1 - 4 - 7 - 10 ... --> 50/3 = 16 + 1 = 17

One more C

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11 Apr 2007, 05:38
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