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How many integers from 0 to 50 inclusive have a remainder of

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Manager
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How many integers from 0 to 50 inclusive have a remainder of [#permalink] New post 12 May 2008, 07:46
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
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Re: integer [#permalink] New post 12 May 2008, 07:51
17

1, 4, 7, 10 .....
but there has to be an easier way
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Re: integer [#permalink] New post 12 May 2008, 08:39
puma 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


ok N=3K+1 if K=0..then N=1 is the smallest possible value..

now..49/3 gives remainder 1..

so
49=1+(n-1)3
48=3n-3
51=3n
n=17...
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Re: integer [#permalink] New post 12 May 2008, 08:45
fresinha12 wrote:
puma 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


ok N=3K+1 if K=0..then N=1 is the smallest possible value..

now..49/3 gives remainder 1..

so
49=1+(n-1)3
48=3n-3
51=3n
n=17...


it is amazing that you managed to apply the arithmetic progression formula to this question.
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Re: integer [#permalink] New post 12 May 2008, 08:48
When I write them all out, I get B, 16.

How can 1 have a remainer of 1 when divided by 3?
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Re: integer [#permalink] New post 12 May 2008, 08:56
RyanDe680 wrote:
When I write them all out, I get B, 16.

How can 1 have a remainer of 1 when divided by 3?


1 divided by 3 is zero with a remainder of 1
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Re: integer [#permalink] New post 12 May 2008, 08:56
it should be 16.

3*16 = 48 ,so only 16 possible numbers.
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Re: integer [#permalink] New post 12 May 2008, 08:58
interesting consideration gmatnub
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Re: integer [#permalink] New post 12 May 2008, 09:10
hey ..i just got bumped to CEO..wut wut :twisted:

gmatnub wrote:
fresinha12 wrote:
puma 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


ok N=3K+1 if K=0..then N=1 is the smallest possible value..

now..49/3 gives remainder 1..

so
49=1+(n-1)3
48=3n-3
51=3n
n=17...


it is amazing that you managed to apply the arithmetic progression formula to this question.
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Re: integer [#permalink] New post 12 May 2008, 09:19
got 17 by using AP series l = a + (n-1)d
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Re: integer [#permalink] New post 12 May 2008, 09:51
OA is C, but I also got B at first
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Re: integer [#permalink] New post 12 May 2008, 12:32
easy way to do this is
add 1 to 50

you get 51, divide this by 3. the answer is 17
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Re: integer [#permalink] New post 12 May 2008, 17:00
Haha .. Nice work .. I am hot on your trail buddy only 800 more posts....

fresinha12 wrote:
hey ..i just got bumped to CEO..wut wut :twisted:
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Re: integer [#permalink] New post 12 May 2008, 18:17
buffdaddy wrote:
easy way to do this is
add 1 to 50

you get 51, divide this by 3. the answer is 17


Wow, great method. Can you provide the reasoning behind this?
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Re: integer [#permalink] New post 13 May 2008, 11:32
Count of numbers between 0-50 (all inclusive) = 51

0,1,2,3,4,5,6,7.....

Numbers which have a remainder of 1 when divided by 3 = 1,4,7,10, etc... one in every 3 numbers!

So total count of numbers which satisfy the condition = 51/3 = 17
Re: integer   [#permalink] 13 May 2008, 11:32
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