I want to ask, if my solution is correct. I am supposed to find the biggest power of number three, which is the biggest divisor of 41*42*43.....250.
My solution: $ \left \lfloor \frac{250}{42} \right \rfloor $ =5
So, the biggest power is five.
Is that correct?
Thanks for any advice.
Let us look at a smaller problem so that we can describe what is going on.
If we wanted to we could actually calculate $10!$ as $3628800$ and factor it as $2^8\cdot 3^4\cdot 5^2\cdot 7$. We see then as a result the largest power of $3$ which divides evenly into $10!$ is $3^4$.
Notice, this has nothing to do with the first term in the product divided by the last term: $\frac{10}{1}$ as your attempt suggested you might have thought.
It is not feasible to try to actually calculate your product and it is not feasible to try to factor your product either, so let us look at how we might approach the problem without needing to do that.
Let us go through and list (or rather, merely count) out all multiples of three in the range in question. $3,6,9$. Each of these will contribute one to the total exponent since they are each a multiple of three. We then continue and further list (or count) each multiple of $9$ in the range as this will have contributed a second factor. We could then also look at the number of multiples of $27$ and $81$ and so on as these contribute yet even more to the total exponent.
In the comments it appears as though you have noticed the intended pattern:
The exponent of the largest power of $3$ which divides evenly into $n!$ is:
$$\left\lfloor\frac{n}{3}\right\rfloor+\left\lfloor\frac{n}{3^2}\right\rfloor+\left\lfloor\frac{n}{3^3}\right\rfloor+\left\lfloor\frac{n}{3^4}\right\rfloor+\dots$$
Now, if we were interested in the product $41\cdot 42\cdot 43\cdots 250$, this is equal to $\frac{250!}{40!}$ so not only should we find the largest power of $3$ which divides $250!$ but we also need to find the largest power of $3$ which divides $40!$ and make a final conclusion.