Question

Five numbers A, B, C, D and E are to be arranged in an array in such a manner that they have a common prime factor between two consecutive numbers. These integers are such that:

A has a prime factor P.
B has two prime factors Q and R.
C has two prime factors Q and S.
D has two prime factors P and S.
E has two prime factors P and R.

If B must be arranged at one end in the array, in how many ways can the other four numbers be arranged?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

B can only sit next to C or E; branch on each choice and follow the forced chain of shared primes to the end, then add up the valid completions.

Answer 1

The Correct Option is C

Step 1: List every pair of numbers that can sit together.
Using the prime factors from the passage: A has P, B has Q and R, C has Q and S, D has P and S, E has P and R.
Two numbers can be neighbours only if they share a prime. Checking every pair: A and D share P, A and E share P, D and E share P, B and C share Q, B and E share R, and C and D share S. These six pairs are the only possible neighbour pairs among all ten pairs of five numbers.

Step 2: Fix B at one end and see what can sit next to it.
B's only possible neighbours are C (through Q) and E (through R). So whichever number sits right next to B must be either C or E. We check both branches completely.

Step 3: Branch where C sits next to B.
Sequence so far: B, C, ...
C's only remaining possible neighbour, other than B which is already used, is D, through S, so the next number must be D. Sequence: B, C, D, ...
D can now be followed by A or E, since both share P with D, and C is already used.
If D is followed by A: the last remaining number is E, and A and E share P, so this closes correctly: B, C, D, A, E. This is a valid full array.
If D is followed by E: the last remaining number is A, and E and A share P, so this also closes correctly: B, C, D, E, A. This is a second valid full array.
So the C-branch gives 2 valid arrangements.

Step 4: Branch where E sits next to B.
Sequence so far: B, E, ...
E can be followed by A or D, since both share P with E, and B is already used.
If E is followed by A: A can then be followed by D, since they share P, and the last number left is C, and D and C share S, so this closes correctly: B, E, A, D, C. This is a third valid full array.
If E is followed by D: D can then be followed by A or C. If D is followed by A, the only number left is C, but A and C share no prime, so this path fails. If D is followed by C, the only number left is A, but C and A share no prime, so this path also fails. So this half of the E-branch gives no valid completion.
So the E-branch gives only 1 valid arrangement.

Step 5: Count the total.
Adding the valid arrangements from both branches: 2 from the C-branch plus 1 from the E-branch equals 3 valid ways to arrange the other four numbers once B is fixed at that end.

Final Answer:
There are 3 valid ways to arrange A, C, D and E once B is pinned at one end.
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