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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
19 Aug 2020, 12:01
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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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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
19 Aug 2020, 12:01
Post A Detailed Correct Solution For The Above Questions And Get A Kudos. Question From Our New Project: GRE Quant Challenge Questions Daily  NEW EDITION!For more theory and practice see our GRE  Math Book
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
20 Aug 2020, 03:53
I did it in a tedious way: The integers that have a remainder of 1 when divided by 3 are of the form 3k+1 where k is an integer. We want to find the highest k such that 3k+1<=50 Now 50=3*16+2, however 49=3*16+1. So the last k that satisfies the condition is k=16. every number of the form 3k+1 with k from 0 to 16 satisfies the condition > there are 17 numbers like this.



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
20 Aug 2020, 05:33
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if a number when divided by 3 and has reminder of 1 then the number is written as 3K+1, so we can consider all the multiples of 3 which are less than 50(because if 50 is considered then 50+1 is 51, which is 51), so numbers will be 0, 3,6...48(which are multiple of 3 less than 50) and last multiple is 48, that is 3*16, so total count is 17(including 0). Hence Answer is C



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
20 Aug 2020, 09:46
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Carcass 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 Remind that the number of terms in an arithmetic sequence is \(\frac{lf}{d} + 1\), where \(l, f\) and \(d\) are its last term, it first term and its common difference, respectively. The numbers between \(0\) and \(50\), inclusive with the remainder \(1\), when divided by \(3\) are \(1, 4, 7, ... , 49\), which is an arithmetic sequence. Thus, the number of terms is \(\frac{491}{3}+1 = \frac{48}{3}+1 = 16+1 = 17\). Therefore, the right answer is C.



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
03 Oct 2020, 06:15
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Carcass 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 Although there are different (and clever) ways to answer this question, I think the fastest (and most accurate) approach is to simply list the values in your head as you count on your fingers (or use a tally sheet) We get: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49 Answer: C Cheers, Brent
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
07 Oct 2020, 18:28
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If you want to solve problems like these under 1 minute then always approach it by dividing the number by the given integer
how to know the pattern 3*1 + 1= 4 (rem = 1when div by 3) 3*2 + 1 = 7 (rem = 1when div by 3) .....
50/3= 16.667 so we know 3*16 = 48 (hence 16 numbers are to be there which will give rem = 1) because 48 + 1 is < 50 finally add 1 because it'll give the same rem so Answer is C >17




Re: How many integers from 0 to 50, inclusive, have a remainder
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07 Oct 2020, 18:28





