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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.

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

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.
_________________

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.

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.

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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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 \(\frac{48-0}{3}+1=17\) numbers divisible by 3 in the range 0-50 inclusive (check this: totally-basic-94862.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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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.

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!

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

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: 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:

50-0+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 ?

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.

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