Correct Option (3)
Let the number of pens received by children A, B, C, D, E, and F be A, B, C, D, E, and F respectively. According to the given conditions:
- 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
The total number of pens, T, is the sum of pens received by all children:
T = A + B + C + D + E + F
Substitute the expressions for B, C, D, E, F in terms of A:
T = A + A/2 + A/3 + A/4 + A/5 + A/6
Factor out A:
T = A (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6)
Find a common denominator for the fractions (LCM of 1, 2, 3, 4, 5, 6 is 60):
T = A (60/60 + 30/60 + 20/60 + 15/60 + 12/60 + 10/60)
T = A ( (60 + 30 + 20 + 15 + 12 + 10) / 60 )
T = A (147/60)
From this, we can express A in terms of T:
A = T * (60/147)
Simplify the fraction 60/147 by dividing both numerator and denominator by their greatest common divisor, which is 3:
A = T * (20/49)
For A to be an integer, T must be a multiple of 49. Among the given options, 147 and 294 are multiples of 49 (147 = 3 * 49, 294 = 6 * 49).
Let's test T = 294:
- A = 294 * (20/49) = 6 * 20 = 120
- B = A/2 = 120/2 = 60
- C = A/3 = 120/3 = 40
- D = A/4 = 120/4 = 30
- E = A/5 = 120/5 = 24
- F = A/6 = 120/6 = 20
In this case, all children (A, B, C, D, E, F) receive an even number of pens (120, 60, 40, 30, 24, 20 respectively). This satisfies the condition that the number of pens each one gets is an even number. Since the question asks for the minimum number of pens, and 147 (as shown below) does not satisfy the condition, 294 is the correct minimum total.
Incorrect Options:
Option 1: 147
If T = 147:
- A = 147 * (20/49) = 3 * 20 = 60
- B = A/2 = 60/2 = 30
- C = A/3 = 60/3 = 20
- D = A/4 = 60/4 = 15
- E = A/5 = 60/5 = 12
- F = A/6 = 60/6 = 10
In this distribution, child D receives 15 pens, which is an odd number. This violates the condition that the number of pens each child gets must be an even number. Therefore, 147 is not the correct answer.
Options 2 and 4: 150 and 300
As derived, for the number of pens A to be an integer, the total number of pens T must be a multiple of 49. Neither 150 nor 300 are multiples of 49. Therefore, if T were 150 or 300, A would not be an integer, which is not possible for a number of pens. Hence, these options are incorrect.