There are between 100 and 110 cards in a collection of cards

Question

There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?

(A) 101
(B) 103
(C) 106
(D) 107
(E) 109

PS20386

Show Answer

Official Answer

A

(A) 101
(B) 103
(C) 106
(D) 107
(E) 109

PS20386

Bunuel · Accepted Answer

If the cards are counted out 3 at a time, there are 2 left over: x=3q+2. From the numbers from 100 to 110 following three give the remainder of 2 upon division by 3: 101, 104 and 107;

If the cards are counted out 4 at a time, there are 1 left over: x=4p+1. From the numbers from 100 to 110 following three give the remainder of 1 upon division by 4: 101, 105 and 109;

Since x, the number of cards, should satisfy both conditions then it equals to 101.

Answer: A.

LalaB · Answer

It is not a combinatorics question. this question is about a remainder

let our number be x

then u are said in the question stem that -

x/3 =y+2 (remainder 2)

x/4=z+1 (remainder 1)

take 101

101/3=33*3+2

101/4=25*4+1

so, it means A is the answer

GMAT40 · Answer

Hi Bunnel,

Is the below solution correct

n = 3q + 2 ==> put values we get 2,  5 , 8

n = 4q + 1 ==> put values we get 1,  5 , 9

LCM is 12

so number should be 12q + 5

101

Ans : A

blueseas · Answer

i think its working. lets take a similar example just to verify: a nymber when divided by 4 gives remainder 2 and when divided by 5 gives remainder 1 LCM of 4 and 5 = 20 now we will calculate first number which is satisfying this condition: when divided by 4 gives remainder 2: 2 6 10 when divided by 5 gives remainder 1: 1 6 11 hence the number in generalized form = 20k+6 lets verify put k=1...number =26==>satisfies both k=2....number = 46==>satisfying both... so thats working. takeaway: step 1: find the LCM step 2: find the first number satisfying both condition ...and then generalise. :-D

mau5 · Answer

In Addition,this sum might help with the concept : positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752.html

jlgdr · Answer

Yes actually the way of solving is getting that X = 12K +5 So then X is a multiple of 12 plus 5. Only number in the given range is 101 Hope this clarifies Cheers! J

Skywalker18 · Answer

Let number of cards = x
If the cards are counted 3 at a time , there are 2 left over-
x= 3p+2
x can take values 101 , 104 , 107

If the cards are counted 4 at a time , there is 1 left over
x= 4q+1
x can take values 101 , 105 , 109

Therefore , x = 101

Answer A

EMPOWERgmatRichC · Answer

Hi All,

We're told that there are between 100 and 110 cards in a collection of cards. If the cards are counted out 3 at a time, then there are 2 left over, but if they are counted out 4 at a time, then there is 1 left over. We're asked for the total number of cards in the collection. This question can be approached in a couple of different ways, including by TESTing THE ANSWERS.

When the cards are dealt out in groups of 3, there are 2 'left over'; this means that the total is "2 more" than a multiple of 3. So, when we subtract 2 from each of these answers, which ones are a multiple of 3...?

Answer A: 101 -2 = 99 --> this IS a multiple of 3 
Answer B: 103 - 2= 101 --> this is NOT a multiple of 3 
Answer C: 106 - 2 = 104 --> this is NOT a multiple of 3 
Answer D: 107 - 2 = 105 --> this IS a multiple of 3 
Answer E: 109 - 2 = 107 --> this is NOT a multiple of 3

With the 2 remaining answers (Answers A and Answer D), we also know that when the cards are dealt in groups of 4, there is 1 'left over'; this means that the total is "1 more" than a multiple of 4. So when we subtract 1 from each of these two answers, which is a multiple of 4...

Answer A: 101 - 1 = 100 --> this IS a multiple of 4 
Answer D: 107 - 1 = 106 --> this is NOT a multiple of 4

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

So first we can find the set of numbers between 100 and 110 which,when divided by 3, have a reminder of 2.

The set includes: 101, 104, and 107.

Next we can find the set of numbers between 100 and 110 which, when divided by 4, have a reminder of 1.

Teh set includes: 101, 105, and 109.

The number common to both sets provides the answer. Thus, there must be 101 cards in the collection.

Alternate Solution:

Let x represent the number of cards in this collection. Since 2 cards are left over when the cards are counted 3 at a time, the remainder when x is divided by 3 is 2 or, in other words, x = 3s + 2 for some integer s. Since one card is left over when the cards are counted 4 at a time, the remainder when x is divided by 4 is 1 or, in other words, x = 4k + 1 for some integer k.

Notice that if we add 7 to x, we obtain x + 7 = 3s + 9 = 4k + 8. Notice also that x + 7 is divisible by both 3 and 4; therefore it is divisible by 12. Thus, the smallest possible value of x + 7 is 12; which means the smallest possible value of x is 5. Every other possible value of x can be found by adding multiples of 12 to 5; such as 17, 29, 41 etc. If we add 12 x 8 = 96 to 5, we obtain 96 + 5 = 101.

Answer: A

Saiharish98 · Answer

Given conditions are 3a+2;4k+1 a, k are integers.. 
Go with options 103,106,107 leave remainder 3 when divided with 4 so straight away eliminate them. 
We are left with 101,109...101 leaves remainder 2 when divided by 3.
Answer.. A

Posted from my mobile device

Nad13091995 · Answer

Hit and trial ..
If the card be 101 then and only then after dividing it by 3 and 4 gives remainder 2 and 1.
Therefore option A is perfectly suited

Posted from my mobile device

fdomeyko · Answer

I follow a different approach.

The number we are looking for (N) meets to conditions:

1) N = 3x + 2
2) N = 4y + 1

For the second condition, we know that the number is odd, since 4y is Even, and Even + 1 = Odd.

So we left with:

101
103
105
107
109

Them we have to apply the divisibility rule of 3 for each of those numbers when substracting 2 from them, since N = 3x + 2, which leave us with the only option possible of 101. (101-2 = 99, and 99 = 3 x 33).

danielpincente · Answer

I am so confused - when counting by 3's, why would you start with 101 and not 100?
Counting by 3's from a range of 100-110 is:  100, 101, 102  then  103, 104, 105,  then  106, 107, 108,  with two left over - 109 and 110

I don't understand how the possible numbers of 101, 104, 107 are determined. Can someone please help?

Bunuel · Answer

I think you are missing a point.

When cards are counted out 3 at a time, there are 2 left over, implies that the number of cards is 2 more than a multiple of 3. Given that there are between 100 and 110 cards, then the number of cards can be 101, 104 and 107 (only these number give a remainder of 2 when divided by 3 in the range from 100 to 110).

When cards are counted out 4 at a time, there is 1 left over, implies that the number of cards is 1 more than a multiple of 4. Given that there are between 100 and 110 cards, then the number of cards can be 101, 105 and 109 (only these number give a remainder of 1 when divided by 4 in the range from 100 to 110).

Thus, the number of cards can only be 101 (only 101 is both 2 more than a multiple of 3 and 1 more than a multiple of 4 in the range from 100 to 110).

Hope it's clear.

DanTheGMATMan · Answer

Standard remainder problem. Find where these criteria coincide:

https://www.youtube.com/watch?v=lPKv08rXzQ4

BrushMyQuant · Answer

Method 1 : Logic

100 is divisible by 4 
=> 101 will give 1 remainder when divided by 4

Remainder of 101 by 3 = Remainder of 1 + 0 + = 2 by 3 = 2
=> 101 satisfies both the conditions!

Method 2 : Algebra

There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over

Dividend = Divisor * Quotient + Remainder 
=> Number of cards, n = 3*a + 2 (where a is the quotient)

but if they are counted out 4 at a time, there is 1 left over.  
=> n = 4*b + 1 (where b is the quotient)

=> n = 3a +2 = 4b + 1
=> 4b = 3a + 1
=> b =  \frac{3a + 1}{4}

Now, only those values of a which make b also an integer all possible values 
=> a = 33 (as 33*3 + 2 = 101 will be between 100 and 110)
=> b =  \frac{3 * 33 + 1}{4}  = 25 => POSSBILE

So,  n = 101

So,  Answer will be A 
Hope it helps!

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

Komal324 · Answer

Hi, Bunuel,

Thank you for sharing an easy solution for the question. I just wanted to check with you that 110/3 also gives a remainder 2. However, we did not consider it because the question says between 100 and 110, that means excluding 100 and 110. Am I correct?

Best,
Komal

Bunuel · Answer

Yes, you're correct. The phrase "between 100 and 110" means greater than 100 and less than 110, so both 100 and 110 are excluded. That’s why 110 isn’t considered, even though it satisfies the remainder condition.

There are between 100 and 110 cards in a collection of cards

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

There are between 100 and 110 cards in a collection of cards

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