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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
 24 Mar 2009, 23:51
Question Stats:
62% (01:36) correct
38% (00:46) wrong based on 100 sessions
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
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]
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 [#permalink]
25 Mar 2009, 01:10
The range of numbers that leave a remainder of 1:
1-49
Therefore total number of integers:
(49-1)/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]
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+(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
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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=a2-a1=3 an-I know that 51/17=3, in order to have a reminder +1, i will take x=16 an=16*3+1=49 N=[(an-a1)/d]+1= (49-4/3)+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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bibha wrote: Hey pkit,
an-I 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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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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Raghav,
Your logic is good but how did you find out there are 16 numbers? If one does it manually, won't it be time consuming?
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How many integers from 0 to 50, inclusive, have a remainder [#permalink]
14 Jun 2010, 01:11
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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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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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Good to know something new every time i login on this forum Thanks
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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 -> (49-1)/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]
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]
31 Aug 2010, 23:56
I think we can find this very quickly by using this:
approximately 1/3 of the numbers from 0-50 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 non-zero numbers though
don't forget to add one..before you are done!
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Re: PS: 0 to 50 inclusive, remainder [#permalink]
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]
06 Jan 2012, 06:22
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Re: PS: 0 to 50 inclusive, remainder [#permalink]
06 Jan 2012, 20:57
First, lets look at the range. With 1 remainder, 4-49. So, number of elements = (49-3)/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]
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]
05 Nov 2012, 07:18
Hi guys, I understand the solution, but I am not able to understand why the Answer shows D --> 18?? Bunuel, please help
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Re: How many integers from 0 to 50, inclusive, have a remainder
[#permalink]
05 Nov 2012, 07:18
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