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

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How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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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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28 Jan 2006, 11:34
50/3= 16,...

The number infront of the comma is the answer; never round up!!!!
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28 Jan 2006, 11:35
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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28 Jan 2006, 11:40
Senior Manager
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28 Jan 2006, 15:54
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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28 Jan 2006, 19:02
TeHCM wrote:
3X + 1 = 50

x = 16.3333..

16 it is

I forgot to count 1

17 it is
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28 Jan 2006, 22:08
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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28 Jan 2006, 22:17
Well, the OA is 17.

3x+1 = 50

x = 16.333...

need to count a '1', so answer is 17.
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16 Mar 2006, 17:43
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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16 Mar 2006, 18:30
Good question buckkitty... can somebody please clarify, step-by-step?
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16 Mar 2006, 20:02
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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16 Mar 2006, 22:20
Remainder 1:
1,4,7,10 ... 49

Number of integers = (49-1/3) + 1 = 17
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17 Mar 2006, 07:13
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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17 Mar 2006, 07:19
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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17 Mar 2006, 11:49
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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17 Mar 2006, 12:22
Can somebody explain
the logic of 3x+1 = 50??
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17 Mar 2006, 20:40
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
17 Mar 2006, 20:40
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# How many integers from 0 to 50, inclusive, have a remainder

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