Find the highest power of two in the expression. What would be the highest power of two in the given expression?
$32!+33!+34!+35!+...+87!+88!+89!+90!\ ?$
I know there are 59 terms involved. I also know the powers of two in each term. I found that $32!$ has 31 two's. If we take 32! out of every term the resulting 59 terms has 2 odd terms and 57 even terms. So it is an even number of the form $2K$. So min possible highest power of 2 will be 32. But I don't know how to calculate the exact value. Surely, We cannot go term by term.
Can anyone throw light in this matter?

You can proceed this way:
$$32!\left(1+33+34\cdot 33+35\cdot 34\cdot 33+\dots+\frac {90!}{32!}\right)$$
$32!$ has factor $2^{31}$, and $1+33$ is of the form $4k+2$, and both $34\cdot 33$ and $35\cdot 34\cdot 33$ are of the form $4k+2$, and for all the terms $36!\over 32!$ and greater we have a divisor $2^2$ or greater.  The three $4k+2$ sums work out as a total sum of the form $4k+2$ and therefore dividing the whole sum by $2^{32}$ will produce an odd number, thus we can conclude that $2^{32}$ is the maximum power of $2$ which divides the specified sum.
