Sally has five red cards numbered 1 through 5 and four blue

Question

Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12

A. 8
B. 9
C. 10
D. 11
E. 12

Bunuel · Accepted Answer

Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12

A) 8
B) 9
C) 10
D) 11
E) 12

The cards are stacked so that the colors alternate:  R - B - R - B - R - B - R - B - R .

We are also told that the number  on each  red card divides evenly into the number on  each neighboring  blue card.

There are two primes in Blue cards: 3 and 5. Their divisors are 1 and 3 AND 1 and 5, respectively. Thus R1 must be between R3 and R5, R3 must be by B3 and R5 must be by B5:  3 - 3 - 1 - 5 - 5 .

Next, add multiple of 3 to the left of  3 :  6 - 3 - 3 - 1 - 5 - 5 .

Add factor of 6 to the left of  6 :  2 - 6 - 3 - 3 - 1 - 5 - 5 .

Add multiple of 2 to the left of  2 :  4 - 2 - 6 - 3 - 3 - 1 - 5 - 5 .

And finally add factor of 4 to the left of  4 :  4 - 4 - 2 - 6 - 3 - 3 - 1 - 5 - 5 .

The sum of the numbers on the middle three cards is  6 + 3 + 3 =12.

Answer: E.

Hope it's clear.

ishdeep18 · Answer

Hi Bunuel, Does such questions actualy appear on GMAT..?? its really tough and makes my mind wooof..!!

Bunuel · Answer

Yes, the question is hard, though I think you can expect such questions if you are doing well on the test.

riteshgmat · Answer

Hi bunuel,

I am not able to understand the logic behind the sequence.
Why it can't be like this
  2 - 6 - 3 - 3 - 4 - 4 - 1 - 5 - 5

And have sum as 3+4+4 = 11

Or any other combination

Bunuel · Answer

Your sequence is not possible because it violates the condition that the number  on each  red card divides evenly into the number on  each neighboring  blue card.

2 - 6 - 3 - 3 -  4  - 4 - 1 - 5 - 5  -->   4   there is NOT a factor of neighboring  3 .

While in the following sequence  4 - 4 - 2 - 6 - 3 - 3 - 1 - 5 - 5  the number on EACH red card is a factor of EACH neighboring blue card.

Hope it's clear.

justbequiet · Answer

1,2,3,4,5 
 3,4,5,6

Rule states that the red card must be a multiple of the 2 blue cards next to it. Therefore, we should start off with the most multiple (6)
 2 - 6 - 3

1, 2,3 ,4,5 
 3,4,5 ,6  
We should then attempt to connect 2 & 3 with a multiple among the remaining red cards
 4 - 2 - 6 - 3 - 3

1, 2,3 ,4,5 
  3,4 ,5, 6  
We then need to look for a factor of 4 & 3.
 4 - 4 - 2 - 6 - 3 - 3 - 1

1,2,3,4 ,5 
  3,4 ,5, 6  
Fill in missing Blue 5 & Red 5 on the far right side
 4 - 4 - 2 - 6 - 3 - 3 - 1 - 5 - 5

We are looking for the sum of the 3 middle digits.
 4 - 4 - 2 -  6 - 3 - 3  - 1 - 5 - 5

6+3+3=12

Answer is E

strawhat316 · Answer

Why can't it be 3-3-6-2-4-4-1-5-5

Sum here equals 10.

Bunuel · Answer

You should read a question and solution more carefully.

Your sequence is not possible because it violates the condition that the number on each red card divides evenly into the number on each neighboring blue card.

3 -3-  6  -2-  4  -4-  1  -5- 5  --> 6, 4, and 1 are NOT factor of neighbouring cards.

strawhat316 · Answer

Got it. I got confused about the order. Thanks Bunuel

Sent from my ONE A2003 using  GMAT Club Forum mobile app

KarishmaB · Answer

Since we have 5 Red and 4 Blue cards, when we alternate them, we will start with the Red card and end with a Red card. 
Each Blue card will be surrounded by 2 Red cards. 2 Red cards will be next to 1 Blue card each while 3 Red cards will be next to 2 Blue cards each.

Blue 3 will have Reds 1 and 3 next to it
Blue 5 will have Reds 1 and 5 next to it
(the prime numbers will only have 1 and themselves next to them) 
Blue 4 will have Reds 2 and 4 next to it
Blue 6 will have Reds 2 and 3 next to it

So we will get the sequence: R5, B5, R1, B3, R3, B6, R2, B4, R4
Note that we could begin with R4 and end with R5 too. R4 and R5 are the only ones used only once so they would be the top and bottom card of the pile.

In any case, the 3 numbers in the middle will be 3, 3 and 6 i.e. total 12.

Answer (E)

Kinshook · Answer

fozzzy

Sally has five red cards numbered 1 through 5 and four blue cards numbered 3 through 6. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards??

A) 8
B) 9
C) 10
D) 11
E) 12

What is the sum of the numbers on the middle three cards?

Red cards = { 1,2,3,4,5 }
Blue cards = { 3,4,5,6 }

Arrangement of cards : - 
 R-  B  -R-  B  -R-  B  -R-  B  -R

4  -4  -  2  -  6  -  3  -  3  -  1  -  5  -  5

The sum of the numbers on the middle 3 cards = 6 + 3 + 3 = 12

IMO E

mkeshri185 · Answer

how do we know which neighbor we have to consider?

Bunuel · Answer

The sequence forces itself step by step. Once you place the first few valid pairs, the neighbors are automatically determined by the divisibility rule. Just follow the logic and the correct neighbors appear on their own.

Sally has five red cards numbered 1 through 5 and four blue

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Sally has five red cards numbered 1 through 5 and four blue

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