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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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24 Mar 2009, 23:51
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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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Last edited by Bunuel on 06 Nov 2012, 03:52, edited 2 times in total.
Renamed the topic and edited the question.



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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25 Mar 2009, 00:16
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C:17
I brute forced this one. All the multiples of 3 + 1 will have a remainder of 1: 4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49  16 numbers total But then I thought, no way it's this easy and thought about 1. 1/3 would also have a remainder of 1, making the answer 17.



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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25 Mar 2009, 01:10
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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



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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25 Mar 2009, 01:21
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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



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Re: Help!!! [#permalink]
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15 May 2010, 22:01
bibha wrote: How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3? A.14 B.15. C.16 D.17 E.18 Ans: C y=x*q+n a1=1*3+1=4, a2=2*3+1=7, => d=a2a1=3 anI know that 51/17=3, in order to have a reminder +1, i will take x=16 an=16*3+1=49 N=[(ana1)/d]+1= (494/3)+1=16 p.s. I may be wrong, if q=0 also could be used than a1=1 and not 4, and than N=17, Ans D. What do you think about this? It is possible or not?
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Last edited by PTK on 16 May 2010, 01:02, edited 2 times in total.



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Re: Help!!! [#permalink]
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16 May 2010, 02:42
bibha wrote: Hey pkit,
anI know that 51/17=3, in order to have a reminder +1, i will take x=16 an=16*3+1=49
Could you please explain this in detail?? I have meant that 51=3*x, than x=17, since we have limits from 0 to 50 inclusive the last number in sequence is 49 which is =16*3+1, as 50=16*3+2.
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Re: Help!!! [#permalink]
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17 May 2010, 09:02
well Bibha, I would like to expalin this in a logical manner rather than by mathematical expression. each no. which is 1 more than multiple of three would be in this series. so upto 50 there are 16 such no.s so correspondingly there are 16 no.s which are 1 more than a multiple of three and the last being 49. reply me if it is not clear.



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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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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 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.
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Re: Help!!! [#permalink]
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14 Jun 2010, 21:53
Does it mean that if zero is included in any similar problem like this then we should consider it, no matter what the divisor is??? Ex if question say any no. between 0 and 50, inclusive, divisible by 3 Answer will still be 17 Is it right
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Re: Help!!! [#permalink]
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Good to know something new every time i login on this forum Thanks
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Re: Help!!! [#permalink]
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15 Jun 2010, 11:54
bibha wrote: How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3? A.14 B.15. C.16 D.17 E.18 ((Last  First)/ n) +1 > (491)/3 +1 = 17 We use 49 because that is the last that will produce a remainder of 1 when divided by 3 and 1/3 has a remainder of 1.



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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22 Jul 2010, 21:13
My answer is C.
There are 48/3=16 numbers that are divisible by 3 (0 is excluded). If we add 1 to each and every of these number, we still have 16 numbers that have a remainder of 1 when divided by 3 (say 4, 7, 10,..., 49). But we have not counted 0 yet, 0+1 equals 1, 1 has a remainder of 1 when divided by 3.
Hence C.



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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31 Aug 2010, 23:56
I think we can find this very quickly by using this: approximately 1/3 of the numbers from 050 are div by 3 so, (50/3) +1 = 16 + 1 = 17 approximately 1/2 of the numbers are div by 2 => (50/2 ) + 1 = 26 approximately 1/4 of the numbers are div by 4 => (50/4) + 1 =13 approximately 1/5 of the numbers are div by 5 => (50/5) + 1 =11 approximately 1/6 of the numbers are div by 6 => (50/6) + 1 = 9 I saw this on Bunuel's post somewhere and I thought to myself, how did he know? so I tried to calculate it and voila!, it is true...didn't try with range starting with nonzero numbers though don't forget to add one..before you are done!



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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01 Sep 2010, 00:12
when 1 is devided by 3 remainder is 1, and when 3 is devided remainder 0. 16*3=48. so upto 48 there will be 16 number which will give remainder 1 after 48, there is only 1 number up to 50 which will give 1 remainder so the answer 17



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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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06 Jan 2012, 20:57
First, lets look at the range. With 1 remainder, 449. So, number of elements = (493)/3 + 1 = 16 Now, lets not forget 1, since 1/3 > 1 as remainder (Good one!) So, 16 +1 = 17
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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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18 Oct 2012, 20:57
General Formula : n= px+q p=3 q=1 n=3x+1 Substituting Values n=0,4,7...49 . x = 0 to 16, inclusive Hence Total nos : 17



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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04 Feb 2013, 23:48
Bunuel wrote: 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.
hi, i was wondering how we got 17 from the calculation that \(p\leq{16\frac{1}{3}}\) ? my approach was: 500+1 = 51 integers total 0/3 has r = 0 1/3 has r = 1 2/3 has r = 2 3/3 has r = 0 ... and so on thus there will be 1 value for every three that will have a remainder of 1 when divided by 3 (cyclicity of 3?) so 51/3 = 17 would this method work for similar questions? i was wondering how we treat the \(p\leq{16\frac{1}{3}}\) term when trying the algebraic method? i.e. how do we know to arrive at 17 ? thanks!



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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink]
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05 Feb 2013, 04:00
essarr wrote: Bunuel wrote: 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.
hi, i was wondering how we got 17 from the calculation that \(p\leq{16\frac{1}{3}}\) ? i was wondering how we treat the \(p\leq{16\frac{1}{3}}\) term when trying the algebraic method? i.e. how do we know to arrive at 17 ? thanks! We have that \(n=3p+1\), where \(p\) is an integer \(\geq{0}\). So, n can be 1 (for p=0), 4 (for p=1), 7, 10, ..., and 49 (for p=16) > 17 values for p > 17 values for n. OR: \(p\leq{16\frac{1}{3}}\) implies that p can take integer values from 0 to 16, inclusive, thus it can take total of 17 values. Hope it's clear.
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Re: How many integers from 0 to 50, inclusive, have a remainder
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