We are given the following information.
$$N = Σ_{i=1}^9n_i$$
$$n_1 = n_2 + n_3$$
$$n_2 = n_4 + n_5$$
$$n_3 = n_6 + n_7$$
$$n_4 = n_8 + n_9$$
$n_i ≥ 0$ for all $\mathbb {Z} $
From this information determine which of the following statements is always true;
$1. n_1 + n_2 + n_3 > n_4 + n_5 + n_6 + n_7 + n_8 + n_9$
$2. n_2 ≥ n_3$
$3. n_1n_6 ≥ n_2n_3$
$4.n_3 ≤ n_2 + n_4 + n_6 + n_8$
$5.\frac{1}{4}N ≤ n_1 ≤ \frac{1}{3}N$
$6.N≠3k+2$ for all integer $k≥0$
$7. ⌊\frac{1}{9}N⌋≤n_5≤⌈\frac{1}{9}N⌉ $
$8.n_8≥n_9$
$9.\frac{1}{3}N≤n_1+n_2≤\frac{1}{2}N$
$10.n_9≤⌈\frac{1}{9}N⌉ $
I have got this far with each statement;
$1.$ because it can be rearranged to $n_1>n_4$ and it is possible for $n_1=n_2=n_4$
$2.$ is straightforward in that $n_2=0$ and therefore $n_1=n_3$ making it untrue
$3.$ I rearranged this so that $n_3(n_3-n_7)≥0$ and because $n_3≥n_7$ this must be always true
$4.$ so it is possible $n_2=n_4=n_6=n_8=0$ and $n_1=n_3=n_7>0$ and therefore this statement is not true
$5.$ $N=3n_1+n_4 ∴ 9n_1+3n_4≤12n_1≤12n_1+4n_4$ which must always be true as $n_1≥n_4$
$6.$ this one doesn't appear to be true, we can let $n_4=2$ and $k=n_1$ and we would get $N=3k+2$ because $N=3n_1+n_4$
$7.$ Not sure where to go with this statement.
$8.$ once again this is just the same as statement $2.$
$9.$ if $n_1=n_2$ and $n_2=n_5$ we could rearrange giving $6n_1+2n_4≤12n_1≤9n_1+3n_4$ disproving the inequaltity
$10.$ Also not sure where to go here.
Apparently only one statement is correct but I have found statements 3 and 5 to always be true. Can someone explain my error and explain how to show 7 and 10 to be incorrect.
For 7: Set $n_6=n_7=n_8=n_9=1$ and $n_5 = 0$
For 10: Set $n_5=n_6=n_7=n_8=0$ and $n_9 = 1$
For 3: Set $n_5=n_7=1$ and $n_6=n_8=n_9=0$