A person X wants to distribute some pens among six children A B C D E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?

Let A, B, C, D, E, F be the number of pens received by each child. From the problem statement, we have the following relationships: A = 2B => B = A/2 A = 3C => C = A/3 A = 4D => D = A/4 A = 5E => E = A/5 A = 6F => F = A/6

For each child to receive an integer number of pens, A must be a multiple of 2, 3, 4, 5, and 6. Therefore, A must be a multiple of the Least Common Multiple (LCM) of these numbers. LCM(2, 3, 4, 5, 6) = 60. So, A can be written as 60k, where k is a positive integer.

Now, let's substitute A = 60k into the expressions for each child's pens: A = 60k B = 60k/2 = 30k C = 60k/3 = 20k D = 60k/4 = 15k E = 60k/5 = 12k F = 60k/6 = 10k

The problem states that the number of pens each one gets must be an even number. Let's check this condition for each child:

1. A = 60k: This is always even, as 60 is even.
2. B = 30k: This is always even, as 30 is even.
3. C = 20k: This is always even, as 20 is even.
4. D = 15k: For 15k to be an even number, k must be an even number (since 15 is odd).
5. E = 12k: This is always even, as 12 is even.
6. F = 10k: This is always even, as 10 is even.

The crucial condition is that D = 15k must be even, which implies that k must be an even integer. To find the minimum number of pens, we choose the smallest positive even integer for k, which is k = 2.

Now, substitute k = 2 back into the expressions for the number of pens: A = 60 * 2 = 120 B = 30 * 2 = 60 C = 20 * 2 = 40 D = 15 * 2 = 30 E = 12 * 2 = 24 F = 10 * 2 = 20

All these numbers (120, 60, 40, 30, 24, 20) are even, satisfying all conditions.

The minimum total number of pens X should buy is the sum of pens for all children: Total = A + B + C + D + E + F Total = 120 + 60 + 40 + 30 + 24 + 20 Total = 294

Analyzing the options: A) 147: Incorrect. This would be the total if A=60 (k=1), but D=15*1=15, which is not even. B) 150: Incorrect. C) 294: Correct, as calculated above. D) 300: Incorrect. This is the total if A=120 and D=30, but it's not the sum of all pens for A=120.

The final answer is C) 294.