# 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've simply stopped calculating one step too soon! You've shown that that expression, with $32!$ divided out is divisible by $2$, but if you just checked whether it was divisible by $4$, you would see that it is indeed not. In particular, the divided expression would be $$\frac{32!}{32!}+\frac{33!}{32!}+\frac{34!}{32!}+\frac{35!}{32!}+\frac{36!}{32!}+\ldots+\frac{90!}{32!}$$ Now, take this mod $4$. All but the first four terms are eliminated, since $4$ divides each of them (as each has $36$ as a factor). However, the first four terms, mod $4$ are $1,\,1,\,2,\,2$, which sums to $2$ mod $4$. Ergo, $2^{33}$ does not divide the original expression, and $2^{32}$ is thus the maximum power of two dividing the expression.
$$32!\left(1+33+34\cdot 33+35\cdot 34\cdot 33+\dots+\frac {90!}{32!}\right)$$
$32!$ 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.
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 $32$.