So I want to divide 100 pens among 5 children in a way that each of them gets an odd number of pens. I think it's not possible but what is the answer and why?
Tag's may be wrong, please feel free to correct me.
$\endgroup$ 44 Answers
$\begingroup$The sum of 5 odd numbers is odd and cannot be equal to 100 that is even
$\endgroup$ $\begingroup$Hint:
- Odd + Odd = Even
- Even + Odd = Odd
- Even + Even = Even
Using these pieces, is the sum of five odds even or odd?
$\endgroup$ $\begingroup$Assume you distributed $100$ pens among $5$ children, so that everyone got an odd number of pens. But addition of $5$ odd numbers gives you an odd number. Thus, adding them back could not give you $100$, since $100$ is even. Thus, by contradiction, you could not have distributed $100$ pens among $5$ children so that every child got an odd number of pens.
$\endgroup$ 0 $\begingroup$An even number $e$ has remainer $0$ when we divided by $2$. We say,
$$e \equiv 0 \pmod 2$$
Keep in mind that in mod $2$ , $x+2k$ where $k$ is an integer means the same as $x$. More generally, $n \pmod k$ means the remainder when we divide $n$ by $k$ (which is equivalent to) plus or minus some integer multiple of $n$.
An odd number on the other hand has remainder $1$:
$$o \equiv 1 \pmod 2$$
There is a rule which say that if $x \equiv a \pmod c$ and $y \equiv b \pmod c$ then:
$$x+y \equiv a+b \pmod c$$
Which is not hard to prove. Therefore we have that the sum of $5$ odd numbers in modulo $2$ is:
$$1+1+1+1+1 \equiv 5 \equiv 1+2(2) \equiv 1 \pmod 2$$
Thus the sum of $5$ odd numbers is always an odd number.
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