How many positive integers less than 100 have a remainder of 2 when

Question

How many positive integers less than 100 have a remainder of 2 when divided by 13?

A. 6
B. 7
C. 8
D. 9
E. 10

15, 28,41,54,67,80,93 = 7
but the set includes 2 which makes answer choice C right ? why 2 is included in the set ??

ankitranjan · Accepted Answer

We have to include 2 also.As 13*0 + 2 =2

If somebody says to divide 2 by 13 ,we will be telling we have 0 quotient and remainder as 2.

ScottTargetTestPrep · Answer

The first positive integer that has a remainder of 2 when divided by 13 is 2. The greatest positive integer less than 100 that has a remainder of 2 when divided by 13 is 93.

So, the total number of numbers that have a remainder of 2 when divided by 13 is (93 - 2)/13 + 1 = 7 + 1 = 8.

Answer: C

IanStewart · Answer

I've mentioned it before on this forum, but the answer to the question above is hopefully clear if you remember how you first learned quotients and remainders. If you have 23 apples and 6 children, and you want to give each child the same number of apples, you can give each child at most 3 apples, and then you'll have 5 apples left over. So when we divide 23 by 6, the quotient is 3 and the remainder is 5.

If instead you have 2 apples and 13 children, then you can give each child 0 apples, and you have 2 left over. So when you divide 2 by 13, the quotient is 0 and the remainder is 2.

If instead you have 2 apples and 13 children, then you can give each child 0 apples, and you have 2 left over. So when you divide 2 by 13, the quotient is 0 and the remainder is 2.

shrive555 · Answer

i'll remember apples and children

FB2017 · Answer

Highest multiple of 13<100 is 91 which is of the form 13*7, hence we have 7 multiples plus 2 in each that leaves a remainder of 2 when divided by 13. Also when 13 divides 2 it leaves a remainder of 2. In total we have 8 integers<100 in this situation.
Please advise if the approach is relevant.

gracie · Answer

2+13x<100
13x<98
x<7.5
x=7
1+7=8
C

CrackverbalGMAT · Answer

Hello Shrive555,

Whenever you deal with  questions related to divisibility , remember that the  basic algorithm  at work is the one you see below:

Dividend = Divisor * Quotient + Remainder.

A couple of points to note when dealing with an equation like the above:
1) Remember that  all the quantities in the equation above ARE TO BE integers .
2) Remember that  any of the quantities (including the quotient) can be ZERO.

Let’s take an example. If there are 28 chocolates which are to be distributed among 13 children in such a way that each child gets the same number of chocos (else, who would want to deal with all the wailing ? ), can this be done?? A very common and quick answer would be NO. I’d say that is wrong. You can still divide 28 by 13 in such a way that each child gets 2 chocolates, but you would be left with 2 at the end.

You will quickly realise that 28 represents the dividend, 13 the divisor, 2 the quotient and the other 2, the remainder. In this case,  the dividend was greater than the divisor.

In another case, if you had only 2 chocolates with you and there were 13 children standing in front of you and you had to distribute the chocolates in such a way that every child got same number of chocolates (breaking of a chocolate into pieces not allowed unless you want to set off a wave of wailing and arm flailing), could it be done? The answer would be NO, but I would argue that it can be done – every child gets ZERO chocolates (i.e. no chocolate) and you’d be left with the 2 that you originally had. Sounds a little counter intuitive, but is true.

Now, here, the dividend was 2 whereas the divisor was 13; the quotient came out to be 0 and the remainder was 2. In this case, the dividend was smaller than the divisor.  When the dividend is smaller than the divisor, the dividend itself is the remainder.

I hope this clarifies why 2 is included in the list.

There are a total of 7 multiples of 13 which are less than 100. Adding the remainder of 2 to these multiples, we obtain 7 numbers, which when divided by 13 give a remainder of 2. 
With the inclusion of 2, the list has 8 numbers i.e. 2, 15, 28, 41, 53, 67, 80 and 93.  The correct answer option is C .

Hope that helps!

CrackverbalGMAT · Answer

shrive555 ,

Whenever you solve such questions, always think " How do such numbers look like? What is the form of such numbers? ". In this particular question, the answer is:

"13Q + 2" ( since Dividend = Divisor * Quotient + Remainder. )

where " 13 " is the divisor, " 2 " is the remainder. These two are fixed since these are provided in the question itself & " Q " is a variable. You can put " Q " as  any of the integers . I mean any (0, -1, -2 etc). This equation will not be harmed.

Now the question statement says " How many positive integers ... ", Q can not be negative & thus the minimum value that "Q" can have is "0". If we put q ="0" in the equation "13Q + 2", we get 2. Go on & put Q=1 now & you will get 15 and so on. This becomes a series now & it will look like:

2, 15, 28 & so on...
 
This becomes an arithmetic progression with "13" being the  common difference  & "2" being the  first term . In this question, we need to count the number of integers less than 100, so maybe one can jot down such numbers & count them in an exam. These will be:

2, 15, 28, 41, 54, 67, 80, 93 -> count them & answer is 8. It took me not more than 15 seconds to jot them down. I might do the same in an exam.

The bigger question, what if we need to count such numbers which are less than 1000? Can we count them then? The answer is a clear NO. How to proceed then?

2 is nothing but 2 + 13*0 (1st term)
15 is nothing but 2 + 13*1 (2nd term)
28 is nothing 2 + 13*2 (3rd term)
41 is nothing but 2 + 13*3 (4th term)
.
.
.

The nth term will be nothing but  2 + 13*(N-1) , correct? (Notice in the 1st term, 13 is getting multiplied with "0". In 2nd term, it is getting multiplied with 1 & so on, in Nth term, it will be getting multiplied with N-1)

Since we need to find integers less than 100, my goal is to find out that Nth term which is less than 100 & follows the same equation.

Thus, my equation becomes:

2 + 13*(N-1)< 100.

Solve this & you will get N < (98/13) + 1 i.e. 8.xx (no need to calculate the decimals here).

Thus, the answer is 8.

Try extrapolating the question to less than 1100, 1500, etc. You will become more confident.

Thanks!

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

How many positive integers less than 100 have a remainder of 2 when

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many positive integers less than 100 have a remainder of 2 when

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards