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M04 #10

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Re: M04 #10 [#permalink] New post 05 Aug 2013, 14:32
eventually, is 0 even or it is neither even nor odd?
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Re: M04 #10 [#permalink] New post 11 Aug 2013, 01:14
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gilda wrote:
eventually, is 0 even or it is neither even nor odd?


Zero is an even integer. (Zero is neither positive nor negative).

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Hope it's clear.
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Re: M04 #10 [#permalink] New post 12 Nov 2013, 08:41
Completely forgot that0!=1
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Re: M04 #10 [#permalink] New post 15 May 2014, 06:58
Check out factorial explanation for those who are confused. This is an explanation from someone else that I liked and makes sense.

You cannot reason that x^0 = 1 by thinking of the meaning of powers as
"repeated multiplications" because you cannot multiply x zero times.
Similarly, you cannot reason out 0! just in terms of the meaning of
factorial because you cannot multiply all the numbers from zero down
to 1 to get 1.

Mathematicians *define* x^0 = 1 in order to make the laws of exponents
work even when the exponents can no longer be thought of as repeated
multiplication. For example, (x^3)(x^5) = x^8 because you can add
exponents. In the same way (x^0)(x^2) should be equal to x^2 by
adding exponents. But that means that x^0 must be 1 because when you
multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense
here.

In the same way, when thinking about combinations we can derive a
formula for "the number of ways of choosing k things from a collection
of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a
group of 5 people shakes hands can be computed using n = 5 (five
people) and k = 2 (2 people per handshake) in this formula. (So the
answer is 5!/(2! 3!) = 10).

Now suppose that there are 2 people and "everybody shakes hands with
everybody else." Obviously there is only one handshake. But what
happens if we put n = 2 (2 people) and k = 2 (2 people per handshake)
in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the
value of 0!. The fraction reduces to 1/x, which must equal 1 since
there is only 1 handshake. The only value of 0! that makes sense here
is 0! = 1.

And so we define 0! = 1.
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Re: M04 #10 [#permalink] New post 22 May 2014, 07:06
Statment 1: 0! or 1! insuff
statement 2: 0,2,4,6,8..., insuff

st1 & st2: x= even & x! =1....therefore x=0...
sufficient

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Re: M04 #10   [#permalink] 22 May 2014, 07:06
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