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

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Economist
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How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   Updated on: 07 Jun 2021, 16:37
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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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Official Answer
C

Originally posted by Economist on 24 Mar 2009, 23:51.
Last edited by Bunuel on 07 Jun 2021, 16:37, edited 3 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 [#permalink] New post   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 implies \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ...

\(n=3p+1\leq{50}\);

\(3p\leq{49}\);

\(p\leq{16\frac{1}{3}}\)

Hence, \(p\), can take 17 values from 0 to 16, inclusive.

Answer: C.­
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   25 Mar 2009, 01:10
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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: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   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: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   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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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   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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How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   14 Jun 2010, 22:05
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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, so there are \(\frac{48-0}{3}+1=17\) numbers divisible by 3 in the range 0-50 inclusive (check this: https://gmatclub.com/forum/totally-basic ... ight=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: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   15 Jun 2010, 11:54
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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: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   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 ?

thanks!
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   05 Feb 2013, 04:00
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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 [#permalink] New post   01 Jul 2015, 16:55
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Calculated the same. \(\frac{Last R1 - First R1}{3} +1 = \frac{(49 - 1)}{3} +1 = 16 + 1 = 17\).

Testing the answer choices if another multiple is needed or if there are one too many. This proved that C was the correct answer.

A. 15 * 3 = 45. 50-45 = R5. Too low.
B. 16 * 3 = 48. 50-48 = R2.
C. 17 * 3 = 51. 50-51 = R1, which is in line with what the question asks.
D. 18 * 3 = 54. 54-50= R4. Too high
E. 19 * 3 = 57. 57-50= R7.Too high
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   26 Sep 2017, 16:24
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Economist 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


The first number that has a remainder of 1 when divided by 3 is 1, and the last number is 49.

Thus, the number of integers from 0 to 50 inclusive that have a remainder of 1 when divided by 3 is:

(49 - 1)/3 + 1 = 17

Answer: C
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   03 Oct 2020, 07:16
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Economist 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] New post   03 Oct 2020, 07:58
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The integer of form 3p + 1 inclusive 0 to 50 will satisfy the given question.

The maximum value of 'p' will be 16 and hence number will be 49 and the Minimum value of 'p' will be 0 and hence number will be 1.

=> 16 + 1= 17

Answer C
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   07 Jun 2021, 16:34
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The pattern in which a division by 3 leaves remainder = 0,1,2
As the remainder 1 is coming once in 3 times => 51/3 = 17 are the no. of values that yield 1 as the remainder when divided by 3.
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   10 Aug 2021, 02:36
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Why complicate things

Listing the numbers is a foolproof, quick-and-easy method
And it takes no time at all
Attachments

File comment: Easiest method
20210810_145347.jpg
20210810_145347.jpg [ 1.17 MiB | Viewed 24905 times ]

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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   19 Oct 2022, 12:16
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We need to find How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3y

Theory: Dividend = Divisor*Quotient + Remainder

Number, n -> Dividend
3 -> Divisor
q -> Quotient (Assume)
1 -> Remainder

=> n = 3*q + 1 = 3q + 1

( Watch this video to learn about Basics of Remainders )

Between 0 and 50
If we put q = 0, we will get the starting number as
n = 3*0 + 1 = 1

If we put q = 16, we will get the ending number as
n = 3*16 + 1 = 48 + 1 = 49

(Hint: here itself we can get the number of possible numbers as 16 + 1 = 17)

So, we have the numbers as 1 , 4, 7, ...., 49

This is an Arithmetic Sequence with

First term, a = 1
Common difference, d = 3
Last term, \(T_n\) = 49

=> Number of terms, n = (\(T_n\) - a) / d + 1 = \(\frac{49 - 1}{3}\) + 1 = \(\frac{48}{3}\) + 1 = 17

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Sequence problems

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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   08 Nov 2022, 01:24
Last quotient divisible by 3 below 50 is 48 , which is 16 * 3 . That means there are 16 numbers that give remainder 0 when divided by 3.
Its also 16 numbers (4 - 49) which give remainder 1 when divided by 3.
1/3 also gives remainder 1.
So, 16 numbers (4 to 49)+ 1 number (1) = 17 numbers give remainder 1 when divided by 3.
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Re: How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   02 Feb 2024, 04:10
Could anyone provide similar questions to practice, please? Thank you!
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How many integers from 0 to 50, inclusive, have a remainder [#permalink] New post   02 Feb 2024, 06:41
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Peachesncream wrote:
Could anyone provide similar questions to practice, please? Thank you!


Similar questions to practice:

Hope it helps.
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