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 number B is not in the list and other four numbers are arranged properly, which of the following must be true?

• A. A is arranged at one end in the array.
• B. C is arranged at one end in the array.
• C. D is arranged at one end in the array.
• D. E is arranged at one end in the array.

Hint:

Once B is removed, check how many numbers each of A, C, D, E can sit next to; the one with only a single possible neighbour must sit at an end.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to arrange A, C, D, E (B is left out) in a row so that any two numbers standing next to each other share a common prime factor. We are asked which statement about the end positions must always hold.

Step 2: Key Formula or Approach:
Model each number as a point and draw a line between two points only when those two numbers share a prime factor. This turns the problem into finding a path that visits every point exactly once, using only the drawn lines. A point that has only ONE line coming out of it can only ever be an end of such a path, since a middle point of the path always needs two different connections, one on each side.

Step 3: Detailed Explanation:
List the shared primes among A, C, D, E only, since B is out of the picture:
$A = \{P\}$, $C = \{Q, S\}$, $D = \{P, S\}$, $E = \{P, R\}$.
A and D share P. A and E share P. D and E share P, so A, D, E all connect to each other through P. C and D share S. C and A share nothing, since A only has P. C and E share nothing, since E has P and R, not Q or S.
So the count of connections for each number is: A connects to D and E (2 connections). D connects to A, E and C (3 connections). E connects to A and D (2 connections). C connects to D only (1 connection).
Since C has just 1 connection, C cannot be placed in the middle of the row, because a middle spot needs 2 different neighbours, so C is forced to sit at an end, right beside D.
Listing out the arrangements that work confirms this: C, D, A, E and C, D, E, A (plus their mirror images E, A, D, C and A, E, D, C) are the only valid rows. C sits at an end in every one of them, while D always sits next to C, one step in from that end, never at the end itself.
Testing the option about A: in C, D, A, E, the number A sits third from the left, not at an end, so "A at an end" fails here.
Testing the option about D: D never reaches an end position in any of the four valid rows, so "D at an end" is false.
Testing the option about E: in C, D, A, E, E sits at the right end, but in A, E, D, C, E sits second from the left, so "E at an end" is not always true.
Only "C at an end" survives every valid case.

Step 4: Final Answer:
Because C keeps only one possible partner once B is removed, C must always sit at one end of the row. The correct option is that C is arranged at one end in the array.
\[ \boxed{\text{C is arranged at one end}} \]