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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
 28 Jan 2006, 11:20
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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
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50/3= 16,...
The number infront of the comma is the answer; never round up!!!!
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VP
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Is it C =17?
There are 16 multiples of 3 and hence there are 16 numbers that leave reminder of 1 (Note checked 48+1 = 49 <50)
1 also leaves a remainder 1 when divided by 3!
Hence 16+1 = 17
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Director
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Senior Manager
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is it 16.
first number is 4, the last number is 49, which gives remainder 1 if divided by 3.
49-4 = 45, 45/3 =15
15+1 =16
the number below 4, and 50 is of no use here
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VP
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TeHCM wrote: 3X + 1 = 50
x = 16.3333..
16 it is
I forgot to count 1 
17 it is
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
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GMAT Club Legend
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I see it as an A.P with 4 as the first term.
1,4,7,10,13,16.......
Last number in this series = 49
Therefore total number of numbers = (49-1)/3 + 1 = 17
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SVP
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Well, the OA is 17.
3x+1 = 50
x = 16.333...
need to count a '1', so answer is 17.
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Senior Manager
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sorry, I know this post is dead but I would like to know how 1 divided by 3 has a remainder of 1? I thought the numerator had to be greater than the denom. in order to have an integer plus an integer remainder. What am I missing?
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Manager
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Good question buckkitty... can somebody please clarify, step-by-step? 
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Senior Manager
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I think I may have answered my own question. (1/3) will have a remainder of 1 when the quotient is 0. (x/y)=q+r ......x=y(q) + r , for integers x and y, with the remainder r. Therefore, 1=3(0) + 1. Is this correctly stated?
I always forget about zero and the tricky roles it plays in questions....
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GMAT Club Legend
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Remainder 1:
1,4,7,10 ... 49
Number of integers = (49-1/3) + 1 = 17
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Current Student
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The trick to this question is in the wording. 1 divided by three does have a remainder of 1, but NOT an integer remainder of 1.
Only the geeks at Pearson would brainstorm a trick question like this...
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Senior Manager
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I thought that a remainder had to be an integer, is that an incorrect assumption?
Grrrrrr!!! I hate these tricky questions!
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Manager
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Let me give it a shot (checking my own understanding in the process)...
Formula for Arithmetic Progression:
an = a1 + (n-1)d
where an = nth term; a1 = first term; n = number of terms; d = difference of succesive numbers
using 4 (which leaves a remainder of 1) as the first term and 49 (which leaves a remainder of 1) as the last term, we can solve for n
49 = 4 + (n-1)3 ---> n = 16
HOWEVER, n = 16 when the first term, i.e. the first number leaving a remainder of 1, is 4. But we have to remember that zero is an integer as well. Therefore, the number 1, when divided by 3 leaves a remainder of 1 as well. The expression would look like this:
3x + 1 = 1
3(0) + 1 = 1
Hence, the answer is 16 + 1 = 17
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Manager
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Can somebody explain
the logic of 3x+1 = 50??
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SVP
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Nayan wrote: Can somebody explain
the logic of 3x+1 = 50??
Find 'x' for which above equation is true or appox true
your will get x = 17 (giving answer 51)
In other words, just remember table of three & check which position gives you number below/equal to 50. 3*16 = 48. So, 16 numbers, but table of 3 doesn't include '1'. So include '1', 16+1 = 17.
Now, why tables of 3? Because each element has remainder of 0 when divided by 3? 
This is a trick, each element + 1, will give remainder of 1. This can be used for any such problem 
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