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How many integers from 0 to 50, inclusive, have a remainder
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Updated on: 06 Nov 2012, 03:52
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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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Originally posted by Economist on 24 Mar 2009, 23:51.
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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How many integers from 0 to 50, inclusive, have a remainder
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14 Jun 2010, 01:11
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: PS: 0 to 50 inclusive, remainder
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25 Mar 2009, 01:10
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
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25 Mar 2009, 00:16
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
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25 Mar 2009, 01:21
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!!!
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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!!!
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14 Jun 2010, 22:05
hardnstrong wrote: 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 0 is a multiple of every integer (except zero itself), so there are \(\frac{480}{3}+1=17\) numbers divisible by 3 in the range 050 inclusive (check this: totallybasic94862.html?highlight=range). But in original question 0 is not considered as one of the numbers: the lowest value of n is 1 (for p=0) and the highest value of n is 49 (for p=16), so total of 17 such numbers.
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Re: Help!!!
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14 Jun 2010, 22:40
Good to know something new every time i login on this forum Thanks
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Re: Help!!!
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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
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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
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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
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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
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06 Jan 2012, 06:22
Algebraic solution is economical.
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Re: PS: 0 to 50 inclusive, remainder
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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
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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
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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
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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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06 Feb 2013, 03:24
The range of given numbers is 500+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
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06 May 2013, 05:41
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+(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, I like this solution using the arithmetic progression but in the expression tn=a+(n1)d why n1 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
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01 Jul 2015, 16:06
I chose C
Since 3 has 16 multiples b/t 050 there must be 16 with a remainder of 1. But also \(1/3\) is included in this list. so 16+1=17=C




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