help please, number of factors

**xinzhuang53** · December 28th 2007, 08:39 AM

wat is the highest # that n can be if 90!/2^n is a whole #

**Plato** · December 28th 2007, 08:42 AM

How many even factors are there in $90!$ ?

**xinzhuang53** · December 28th 2007, 08:46 AM

45?

**Isomorphism** · December 28th 2007, 08:46 AM

It is given by
$[\frac{90}{2}] + [\frac{90}{4}] + [\frac{90}{8}] + [\frac{90}{16}] + [\frac{90}{32}] + [\frac{90}{64}] + [\frac{90}{128}] +.......$
That is, 45+22+11+5+2+1+0+..... = 86
so n=86

**xinzhuang53** · December 28th 2007, 08:56 AM

i still dont get it how did u get that equation, and 90/4 is 22.5, and 90/8 is 11.25.

**Plato** · December 28th 2007, 09:04 AM

Originally Posted by xinzhuang53

i still dont get it how did u get that equation, and 90/4 is 22.5, and 90/8 is 11.25.

That is not simple division. It is the floor function or the greatest integer function.
$\left\lfloor {\frac{{90}}{8}} \right\rfloor = 11$

**Isomorphism** · December 28th 2007, 09:04 AM

Originally Posted by xinzhuang53

i still dont get it how did u get that equation, and 90/4 is 22.5, and 90/8 is 11.25.

[] stand for floor function.

The equation comes from counting all even numbers in 90! that is $[\frac{90}2]$
Then counting all multiples of 4 that give an additional multiple of 2 in 90!, that is $[\frac{90}4]$
Then counting all multiples of 8 that give yet another additional multiple of 2 in 90!, that is $[\frac{90}8]$
And so on....

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