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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?

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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.

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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.

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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 $32$.

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