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

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

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New post 06 Feb 2013, 02:24
The range of given numbers is 50-0+1=51
Any number when divided by 3 can give remainders from the list {0,1,2}. This patten will repeat for every 3 numbers. Hence divide the range by number of available remainders. i.e 51/3=17
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Re: PS: 0 to 50 inclusive, remainder [#permalink]

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New post 06 May 2013, 04:41
abhishekik wrote:
My ans is also C.17.

Explanation:

1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on.
Hence we have an arithmetic progression: 1, 4, 7, 10,..... 49, which are in the form 3n+1.
Now we have to find out number of terms.
tn=a+(n-1)d, where tn is the nth term of an AP, a is the first term and d is the common difference.
so, 49 = 1+(n-1)3
or, (n-1)3 = 48
or, n-1 = 16
or, n = 17


Hello,

I like this solution using the arithmetic progression but in the expression tn=a+(n-1)d why n-1 and not n ?

As I know Un= U0 + n r where U0 is the first terme and r is the common difference.

Thx in advance for the lighting !
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 01 Jul 2015, 15:06
I chose C

Since 3 has 16 multiples b/t 0-50 there must be 16 with a remainder of 1. But also \(1/3\) is included in this list. so 16+1=17=C
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 01 Jul 2015, 15:55
Calculated the same. \(\frac{Last R1 - First R1}{3} +1 = \frac{(49 - 1)}{3} +1 = 16 + 1 = 17\).

Testing the answer choices if another multiple is needed or if there are one too many. This proved that C was the correct answer.

A. 15 * 3 = 45. 50-45 = R5. Too low.
B. 16 * 3 = 48. 50-48 = R2.
C. 17 * 3 = 51. 50-51 = R1, which is in line with what the question asks.
D. 18 * 3 = 54. 54-50= R4. Too high
E. 19 * 3 = 57. 57-50= R7.Too high
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 14 Sep 2016, 08:47
Economist 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


First, when a integer is divided by 3, it can have the remainder 0,1 or 2.

There are 50-0+1=51 integers between 0 and 50 inclusive.

Just a quick example:
integer 0: remainder 0
integer 1: remainder 1
integer 2: remainder 2
integer 3: remainder 0
integer 4: remainder 1 and so on. Practicaly, there are 51/3=17 remainders of 0, 17 remainders of 1 and 17 remainders of 2. Answer: C
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 14 Sep 2016, 09:11
Number of integers from 0 to 50 with remainder of 1 when divided by 3 can be found out as follows:
Number of integers divisible by 3 from 0 to 50: 48 = 3 + (n-1) * 3 ; using the formula for nth term of series in arithmetic progression ,where nth term is 48, 1st term is 3 and difference between terms is 3.
Therefore n=16, since adding 1 to each of these numbers will give remainder 1, however, when 1 is divided by 3, it gives remainder of 1, hence we have to include 1, giving the total number of integers as 16+1 =17
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 14 Sep 2016, 10:03
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

let x=number of integers with a remainder of 1 when divided by 3
range is 1-49
1+3(x-1)=49
x=17
C.
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]

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New post 14 Sep 2016, 11:14
Thank you Bunuel..as always...

Bunuel 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

Algebraic way:

Integer have a remainder of 1 when divided by 3 --> \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ...

\(n=3p+1\leq{50}\) --> \(3p\leq{49}\) --> \(p\leq{16\frac{1}{3}}\) --> so \(p\), can take 17 values from 0 to 16, inclusive.

Answer: C.
Re: How many integers from 0 to 50, inclusive, have a remainder   [#permalink] 14 Sep 2016, 11:14

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