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

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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

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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24 May 2014, 05:54

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

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16 Jun 2015, 07:07

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

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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

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

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04 Aug 2016, 06:23

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

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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

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

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

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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.

gmatclubot

Re: How many integers from 0 to 50, inclusive, have a remainder
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14 Sep 2016, 11:14

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