**The poisoned letters:**the only thing you have to go on is the possibility that the poisoner actually wants an audience, or you'd be dead already - so there must be a feasible solution.

If you construct a truth table then you will see that if one of the statements true at a time you get three different scenarios. Now, testing them for validity you find that if the blue envelope truthful then it is safe, the red one is also safe because it is false. The typed envelope is also false, which fits. Unfortunately this doesn't work because only one is supposed to contain the cure. :-/

If only the red one is true then it is poisoned, blue is lying when it claims to be safe, and therefore the typed one is telling the truth - which it shouldn't be doing. This is a contradiction, so we throw out the red envelope as true. :-(

If the typed one is truthful, then the others are false - meaning that the blue is dangerous and the red is safe... but that belies the notion that the verbose one is true. Also contradictory! We have to discard all three scenarios. >:-(

You should be able to get it from there.

**The weighty problem:**there are several key steps to solving this. First is to divide the packages into three groups of four. The second is to recognize that you can label the packages. If the first weighing results in balance, then those eight are all 'reference' weights, and the other four are where your problem lies. Weigh two against reference boxes. Whether they balance or not, you are down to just two dodgy packages compare one versus a reference and you can ascertain the incorrectly packed item. Easy!

The

__harder task__is when they first weighing results in no balance. Observe the packages: the side that descends, label those with an "H" indicating that IF the package is there, it's heavy. Label the others "L" suggesting that they are the cohort of a possible light package. The four that aren't weighed are now references.

The last sneaky trick is to mix H and L packages on each side... and that's where this hint stops. Try it from there.

_______________________________________________________

Solutions now follow, don't read further if you'd still like to solve the final step for each problem.

**What's your poison?**AHA! You realise that the only truthful label all along was the very first one - all others on the envelopes are false. (Yes, it does say that the box was "labeled" - I snuck that in to befuddle you).

Trying it out, we find that the blue envelope is dangerous, the red one is safe, and the typewritten one is a dirty sniveling fibber (as it should be). Open the RED envelope and breathe deep the cure!

**Oh where, oh where has my little**OK, so we have four 'heavies', four 'lights', and four 'references'. Mix them up, thuswise...

*aberrant shipping container*gone...?[H1 H2 L1] versus [L2 H3 H4]

If the left side drops and the right side rises we're down to three boxes (H1 H2 and L2). If the right drops it's also three (H3 H4 and L1). weigh the two Hs against each other: if they're the same, it's the L; if they're different, it's the H that dropped. If those six are all equal, then just compare L3 and L4. You can see how that works.

In all cases you can find the box that's the wrong weight, and in all but one (the one that is never weighed) you can tell if they are too heavy or too light.

**Oh, and the name of the Secretary General of the United Nations 35 years ago?**Whatever it is

__now__. :-)

ND

(Never-ending Deviousness)