Let the time taken by Jack to complete the work be \(x\) days.
Then, the time taken by John is \(2x\) days.
Let the time taken by Jim be \(y\) days.
From the question, it is given that:
The average time taken by Jack and Jim working together is equal to the time taken by John and Jim working together.
So, we write their combined work equations:
Work done by Jack and Jim together in one day is:
\(\frac{1}{x} + \frac{1}{y}\)
Work done by John and Jim together in one day is:
\(\frac{1}{2x} + \frac{1}{y}\)
Now we are told that the **average time of Jack and Jim** is same as the **time taken by Jack, John, and Jim together**:
From the question, the average of Jack and Jim’s time is equal to the harmonic mean:
We can equate:
\(\frac{2xy}{x + y} = \frac{2x}{3}\) [From question]
Cross-multiplying:
\(\frac{2x}{3} = \frac{xy}{x + y}\)
Multiply both sides by \(x + y\):
\(\frac{2x(x + y)}{3} = xy\)
Multiply and simplify:
\(2x^2 + 2xy = 3xy\)
\(2x^2 = 3xy - 2xy = xy\)
Now divide both sides by x:
\(2x = y\)
So, we have established: \( y = 2x \)
Now consider: John takes 3 days more than what all three of them working together take.
Work done by all three in one day:
\(\frac{1}{x} + \frac{1}{2x} + \frac{1}{y}\)
Substitute \(y = 2x\):
\(\frac{1}{x} + \frac{1}{2x} + \frac{1}{2x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}\)
So, time taken by all three working together = \(\frac{x}{2}\) days.
John’s time = \(2x\) days.
Given: John takes 3 days more than all three together.
So,
\(2x - \frac{x}{2} = 3\)
\(\frac{4x - x}{2} = 3 \Rightarrow \frac{3x}{2} = 3\)
Multiply both sides by 2:
\(3x = 6 \Rightarrow x = 2\)
Now substitute back:
Time taken by Jack = \(x = 2\) days
Time taken by Jim = \(y = 2x = 4\) days
∴ Time taken by Jim to complete the work is \(\boxed{4}\) days.
When $10^{100}$ is divided by 7, the remainder is ?