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

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Manager
Joined: 23 Mar 2008
Posts: 218
How many integers from 0 to 50 inclusive have a remainder of [#permalink]

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12 May 2008, 08:46
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?

a)15
b) 16
c)17
d)18
e)19
Director
Joined: 23 Sep 2007
Posts: 782

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12 May 2008, 08:51
17

1, 4, 7, 10 .....
but there has to be an easier way
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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12 May 2008, 09:39
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...
Director
Joined: 23 Sep 2007
Posts: 782

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12 May 2008, 09:45
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.
Manager
Joined: 27 Jun 2007
Posts: 199

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12 May 2008, 09:48
When I write them all out, I get B, 16.

How can 1 have a remainer of 1 when divided by 3?
Director
Joined: 23 Sep 2007
Posts: 782

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12 May 2008, 09:56
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
Senior Manager
Joined: 02 Dec 2007
Posts: 449

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12 May 2008, 09:56
it should be 16.

3*16 = 48 ,so only 16 possible numbers.
Senior Manager
Joined: 02 Dec 2007
Posts: 449

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12 May 2008, 09:58
interesting consideration gmatnub
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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12 May 2008, 10:10
hey ..i just got bumped to CEO..wut wut

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.
Intern
Joined: 29 Apr 2008
Posts: 21

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12 May 2008, 10:19
got 17 by using AP series l = a + (n-1)d
Manager
Joined: 23 Mar 2008
Posts: 218

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12 May 2008, 10:51
OA is C, but I also got B at first
Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10

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12 May 2008, 13:32
easy way to do this is

you get 51, divide this by 3. the answer is 17
CEO
Joined: 17 May 2007
Posts: 2950

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12 May 2008, 18:00

fresinha12 wrote:
hey ..i just got bumped to CEO..wut wut
Manager
Joined: 05 Feb 2007
Posts: 140

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12 May 2008, 19:17
easy way to do this is

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

Wow, great method. Can you provide the reasoning behind this?
Intern
Joined: 06 Aug 2007
Posts: 38

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13 May 2008, 12:32
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, 12:32
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