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?

The sum
$$32!+33!+34!+35!+ \dotsb +87!+88!+89!+90!=32![1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb]$$
The expression
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 0 \pmod{2}$$
but
$$1+33+33\cdot 34+33\cdot 34\cdot 35+\dotsb \equiv 2 \pmod{4}.$$
Thus all the powers of $2$ will come from $32!$ and only one from the bracketted expression.
The highest exponent of $2$ in $32!$ is given by
$$16+8+4+2+1=31.$$
Thus the highest exponent of $2$ will be $32$.
