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: Set up the sharing relation between the five numbers.
Two numbers can sit next to each other in the array only if they share a prime factor. Write down which primes each number carries: A has P, B has Q and R, C has Q and S, D has P and S, E has P and R.
Check every pair of numbers to see if they share a prime. A and D share P, so they can be neighbours. A and E share P too. D and E both carry P, so they can also be neighbours. B and C share Q. B and E share R. C and D share S. No other pair shares a common prime (A and B, A and C, B and D, C and E do not share anything).

Step 2: Remove B and look only at A, C, D, E.
The question drops B from the array, so only the links among A, C, D, E matter now. From Step 1, once B is gone, the surviving links are: A-D, A-E, D-E, and C-D.
Notice that C now has only ONE possible neighbour left, namely D, because the only other link C had was with B, and B is not in the array. Every other number still has at least two possible neighbours: A can sit next to D or E, D can sit next to A, E, or C, and E can sit next to A or D.

Step 3: Work out what a valid array of A, C, D, E can look like.
In an array of four numbers, the two numbers at the ends need only ONE neighbour each, while the two numbers in the middle need TWO neighbours each (one on each side).
Since C has only one possible partner (D), C cannot ever sit in the middle of the array, because a middle position needs two different neighbours and C simply does not have a second number it can share a prime with. So C is forced to sit at one end, right next to D.
Checking this by listing the arrangements that actually work: C-D-A-E and C-D-E-A (and their reverses E-A-D-C and A-E-D-C) are the only valid orders, since A, D, E share P among themselves and D also connects to C. In every single one of these valid orders, C sits at an end and D sits right next to it, never at an end.

Step 4: Rule out the other options.
Option stating A is at one end is not always true: in the order C-D-A-E, A sits in the third position, not at an end.
Option stating D is at one end is never true here: D always sits second from an end, next to C, since D is the only link C has.
Option stating E is at one end is not always true: in the order C-D-A-E, E sits at the right end, but in A-E-D-C, E sits in the second position, not at an end.
Only "C sits at one end" holds in every valid arrangement.

Final Answer:
Once B is removed, C is left with only one possible neighbour (D), so C must always occupy one end of the array.
\[ \boxed{\text{C is arranged at one end in the array}} \]