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

Question

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

Bunuel · Accepted Answer

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.

shkusira · Answer

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

liubhs02 · Answer

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.

abhishekik · Answer

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

hardnstrong · Answer

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 :?:

Is it right :?:

Bunuel · Answer

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.

GMATBLACKBELT720 · Answer

((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.

essarr · Answer

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!

Bunuel · Answer

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.

mejia401 · Answer

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

JeffTargetTestPrep · Answer

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

BrentGMATPrepNow · Answer

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

Ishi2511 · Answer

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.

vv65 · Answer

Why complicate things

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

BrushMyQuant · Answer

We need to find How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 y

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

https://www.youtube.com/watch?v=EfNT2JoE5CM

Peachesncream · Answer

Could anyone provide similar questions to practice, please? Thank you!

Bunuel · Answer

ruis · Answer

Could we use the following reasoning? 1/3 number will have a remainder of 1 hence = 33%*50 = 16,5. And we round up to 17 due to de decimals (33.33%)... Is that right? Thanks

Bunuel · Answer

If the question were:
 
How many integers from 0 to 52, inclusive, have a remainder of 1 when divided by 3 ?

The answer would be 18. However, employing your method you'd do 52*1/3 = 17.33... and rounding this would give 17, not 18 and you'd be 1 off. So, I'd recommend sticking with the approaches given on the previous page.

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

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

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