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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]  22 Mar 2007, 11:19
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

15
16
17
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19
Director
Joined: 12 Jun 2006
Posts: 536
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Kudos [?]: 28 [0], given: 1

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?
Intern
Joined: 01 Mar 2007
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Shortcut [#permalink]  22 Mar 2007, 12:47
this is an arithmetic sequence, a(n) = 3*n +1

where n can have values from 0 to 16
that is, 17 values.
Director
Joined: 14 Jan 2007
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Kudos [?]: 69 [0], given: 0

arithmetic sequence

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

nth term = a+(n-1)d = 49
a=1
d=3
1+(n-1)*3 = 49
n = 17
Senior Manager
Joined: 20 Feb 2007
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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.

VP
Joined: 28 Mar 2006
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Kudos [?]: 19 [0], given: 0

a=1

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

n=17
Manager
Joined: 20 Mar 2007
Posts: 62
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Kudos [?]: 8 [0], given: 0

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?
Intern
Joined: 11 Feb 2007
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i get 16. you cannot include the numbers less than 3. that will not yeild a remainder of 1.
Manager
Joined: 28 Dec 2006
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Kudos [?]: 0 [0], given: 0

of course u can. 1 divided by 3 also has a remainder of 1.
Director
Joined: 30 Nov 2006
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Location: Kuwait
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Kudos [?]: 167 [0], given: 0

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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