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

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

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

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.

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

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

Sum here equals 10.

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.

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

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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)

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