reasoning based question

Question

If 100 apples are to be divided among 25 people,how they can be divided so that none of them gets an even number of apples?

Give $99$ apples to one person and $1/24$th of an apple to each of the others. $99$ is an odd number and $1/24$ is not an integer, hence not an even number, so nobody gets an even number of apples. — Barry Cipra, May 27 '16 at 17:55
@BarryCipra That's cheating, fractions shouldn't be allowed :D — layman, May 27 '16 at 17:57
this is the core of the problem - the formal negation of "not being even" is "being odd or noninteger" :) — user340508, May 27 '16 at 18:09

user340508 · Answer 1 · 2016-05-27T17:59:00.017

2

The sum of any 25 odd numbers is odd, so this is impossible in the case when everybody gets whole apples.

But formally, the condition

none of them gets an even number of apples

allows noninteger numbers of apples, I mean to divide an apple into 2 halves and give halves to different people ($0.5$ or $3.5$ are NOT EVEN numbers).

So for example:

$27$ apples for person 1,

$3.5$ for person 2 and 3,

$3$ apples for all others.

edited May 27 '16 at 17:59

answered May 27 '16 at 17:55

user340508

911

Maybe one person can have zero apples? Or is this technically a even number? – Tim Huijgens May 27 '16 at 17:56
0 is an even number: $\frac{0}{2}=0\in\mathbb{Z}$ – user340508 May 27 '16 at 17:57
@TimHuijgens I think $0$ is an even number since $2$ divides it. – layman May 27 '16 at 17:58
1

@user46944, see http://math.stackexchange.com/questions/15556/is-zero-odd-or-even – Barry Cipra May 27 '16 at 18:01

score 0 · Answer 2 · answered May 27 '16 at 18:07

0

Give 4 apples to each person and ask them to break off in pairs. Ask a person in every pair to pass one of their apples to the other one in the pair such that in each pair, one has 3 apples and the other has 1 apple. (This all works because no where in the question has it asked to divide them equally among the people. Also, this is one of the many ways in which it can be done)

answered May 27 '16 at 18:07

Achintya Mitra

1

So you have 12.5 pairs. – almagest May 27 '16 at 18:20

score 0 · Answer 3 · answered May 27 '16 at 19:01

The number of ways in which you can do that is given by the coefficient of $x^{100}$ in $$(x^1+x^3+x^5+x^7+\ldots)^{25}\tag{1}$$ i.e. by: $$ [x^{100}]\left(\frac{x}{1-x^2}\right)^{25} = [x^{75}]\frac{1}{(1-x^2)^{25}} \tag{2}$$ and since: $$ \frac{1}{(1-z)^{25}}=\sum_{n\geq 0}\binom{24+n}{n}z^n \tag{3}$$ by stars and bars, $$ \frac{1}{(1-x^2)^{25}}=\sum_{n\geq 0}\binom{24+n}{n}x^{2n} \tag{4}$$ and the RHS of $(2)$ simply equals $\color{red}{0}$.

Unsurprisingly, since the sum of $25$ odd numbers is an odd number, while $100$ is an even number.

score 0 · Answer 4 · answered May 27 '16 at 23:33

With $\quad n_{k} \in \left\lbrace 0,1,2,\ldots,49\right\rbrace\quad\mbox{and}\quad k = 1,2,\ldots 25$:

$$ \sum_{k = 1}^{25}\left(2n_{k} + 1\right) = 2\left(\sum_{k = 1}^{25}n_{k}\right) + 25 \not= 100 $$ because the 'left side' is $\color{#f00}{\mbox{an odd number}}$. It's not possible to do that.

reasoning based question

4 Answers4