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

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

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.

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

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

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

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:

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

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

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

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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

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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).

­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?­

­

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.