If $90! = (90)(89)(88)...(2)(1)$, then what is the exponent of the highest power of $2$ which will divide $90!$ ?
How would I apply one of the easiest method from Here?
I need help on applying the link to this question.
I do not understand which one explains my case, and how I can solve using the method.
I would appreciate if someone showed how it is applied to this
On
Hint:
Your intuition isn't that bad. There are indeed $45$ even factors, hence $2^{45}$. But some factors still have other powers of $2$ (like $4=2^2$ or $48=2^4\cdot3$).
Look at what happens if you erase the odd factors and divide all the even ones by $2$...
You obtain the recurrence $$p(n)=\lfloor \frac n2\rfloor+p\left(\lfloor \frac n2\rfloor\right).$$
We know that there are $\lfloor \frac{90}{2}\rfloor$ integers below $90$ that have at least one factor $2$, $\lfloor \frac{90}{4}\rfloor$ numbers that have $2$ factors $2$, etc. Thus, the number of factors $2$ in $90!$ is $$\sum_{k=1}^{\infty}\left\lfloor \frac{90}{2^k}\right\rfloor=86$$ We don't actually have to do this sum up to $\infty$, but only until $6$, since there's no numbers under $90$ with more than $6$ factors. However, this form is more easily generalizable, which is always nice.