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Manager
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How many integers from 0 to 50 inclusive have a remainder of [#permalink]
 12 May 2008, 08:46
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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Director
Joined: 23 Sep 2007
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17
1, 4, 7, 10 .....
but there has to be an easier way
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Current Student
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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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Director
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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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Manager
Joined: 27 Jun 2007
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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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Director
Joined: 23 Sep 2007
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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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Senior Manager
Joined: 02 Dec 2007
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it should be 16.
3*16 = 48 ,so only 16 possible numbers.
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Senior Manager
Joined: 02 Dec 2007
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interesting consideration gmatnub
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Current Student
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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. |
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Intern
Joined: 29 Apr 2008
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got 17 by using AP series l = a + (n-1)d
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Manager
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OA is C, but I also got B at first
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Director
Joined: 14 Oct 2007
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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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CEO
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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 |
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Manager
Joined: 05 Feb 2007
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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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Intern
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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
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