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Manager  B
Status: May The Force Be With Me (D-DAY 15 May 2012)
Joined: 06 Jan 2012
Posts: 193
Location: India
Concentration: General Management, Entrepreneurship
There are between 100 and 110 cards in a collection of cards  [#permalink]

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3
14 00:00

Difficulty:   15% (low)

Question Stats: 82% (01:42) correct 18% (02:08) wrong based on 395 sessions

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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

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GMAT 1: 690 Q47 V38 Re: Number of Cards Between 100 - 110  [#permalink]

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1
1
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
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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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4
boomtangboy wrote:
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

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.

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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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2
1
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
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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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2
GMAT40 wrote:
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. _________________
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Joined: 10 Oct 2012
Posts: 584
Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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GMAT40 wrote:
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

In Addition,this sum might help with the concept : positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752.html
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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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boomtangboy wrote:
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

.................

divide 101 by 3 and then divide 101 by 4 again, we have 2 and 1 as the remainder respectively.
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Concentration: Finance
Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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1
blueseas wrote:
GMAT40 wrote:
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. 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!
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Joined: 03 Oct 2016
Posts: 120
Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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Dividing by 3 & 4 is the quickest way to get the answer. Refer remainders shortcut on math mega thread.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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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

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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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boomtangboy wrote:
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

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.

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Re: There are between 100 and 110 cards in a collection of cards  [#permalink]

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100≤x ≤110

Based on the info we can form two equations and solve.

X/3 = Q +2/3
x=3Q +2 Q = positive quotient

x/4 = Z = 1/4
x=4Z +1 Z = positive quotient

4Z+1 = 3Q+2
4Z=3Q+1
Z = (3Q+1)/4

Since both Z and Q must be positive (as they are the quotients under each division case), we can plug-in each answer choice and if that answer choice is divisible by 4, allowing Z to be an integer, we have a match.

A is the only one that fits.
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+1 Kudos if I have helped you Re: There are between 100 and 110 cards in a collection of cards   [#permalink] 15 Oct 2019, 21:35
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