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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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01 Jul 2015, 16:55
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. 5045 = R5. Too low. B. 16 * 3 = 48. 5048 = R2. C. 17 * 3 = 51. 5051 = R1, which is in line with what the question asks. D. 18 * 3 = 54. 5450= R4. Too high E. 19 * 3 = 57. 5750= R7.Too high



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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14 Sep 2016, 10:11
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 + (n1) * 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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14 Sep 2016, 11:03
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 149 1+3(x1)=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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26 Sep 2017, 16:24
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 The first number that has a remainder of 1 when divided by 3 is 1, and the last number is 49. Thus, the number of integers from 0 to 50 inclusive that have a remainder of 1 when divided by 3 is: (49  1)/3 + 1 = 17 Answer: C
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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01 Apr 2018, 06:04
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+(n1)d, where tn is the nth term of an AP, a is the first term and d is the common difference. so, 49 = 1+(n1)3 or, (n1)3 = 48 or, n1 = 16 or, n = 17 hello there from this \(49 = 1+(n1)3\) I get following: \(49 = 1+3n3\) \(49=3n2\) \(3n=51\) how did you get these value  > or, (n1)3 = 48 or, n1 = 16 or, n = 17



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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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Updated on: 01 Apr 2018, 07:36
mejia401 wrote: 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. 5045 = R5. Too low. B. 16 * 3 = 48. 5048 = R2. C. 17 * 3 = 51. 5051 = R1, which is in line with what the question asks. D. 18 * 3 = 54. 5450= R4. Too high E. 19 * 3 = 57. 5750= R7.Too high hello there how about this pushpitkc many thanks! Isnt it formula of number of multiples that you used .... but if it is formula of number of multiples in X range than isnt it used when we need to find number of terms divided by 3 and not when we need to find how many integers leave remainder 1 as in my post (#7) https://gmatclub.com/forum/arithmeticp ... l#p2035478
Originally posted by dave13 on 01 Apr 2018, 06:13.
Last edited by dave13 on 01 Apr 2018, 07:36, edited 1 time in total.



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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01 Apr 2018, 06:25
dave13 wrote: 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+(n1)d, where tn is the nth term of an AP, a is the first term and d is the common difference. so, 49 = 1+(n1)3 or, (n1)3 = 48 or, n1 = 16 or, n = 17 hello there from this \(49 = 1+(n1)3\) I get following: \(49 = 1+3n3\) \(49=3n2\) \(3n=51\) how did you get these value  > or, (n1)3 = 48 or, n1 = 16 or, n = 17 Hi dave13We have \(49 = 1+(n1)3\) This can be rewritten as \((n1)3 = 49  1 = 48\) Dividing by 3 on both sides, we get \(n1 = \frac{48}{3} = 16\) Therefore, we can arrive at n = 17 Hope this helps you!
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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01 Apr 2018, 10:51
dave13 wrote: mejia401 wrote: 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. 5045 = R5. Too low. B. 16 * 3 = 48. 5048 = R2. C. 17 * 3 = 51. 5051 = R1, which is in line with what the question asks. D. 18 * 3 = 54. 5450= R4. Too high E. 19 * 3 = 57. 5750= R7.Too high hello there how about this pushpitkc many thanks! Isnt it formula of number of multiples that you used .... but if it is formula of number of multiples in X range than isnt it used when we need to find number of terms divided by 3 and not when we need to find how many integers leave remainder 1 as in my post (#7) https://gmatclub.com/forum/arithmeticp ... l#p2035478Hi dave13Yes, the formula only calculates the number of terms divided by 3. But if you observe every(a number divisible by 3 + 1) will give a remainder of 1 when divided by 3 Here, between 0 and 50(both inclusive), there are 17 numbers which are divisible by 3  0(0*3) to 48(16*3) 17 numbers which leave the remainder 1(0*3 + 1)  49(16*3 + 1) Hope this helps clear your confusion!
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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08 Apr 2018, 06:58
The range of numbers that leave a remainder of 1: 149
Therefore total number of integers: (491)/3 + 1 = 17
The last +1 is because the question is inclusive.
Hence C




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