(Missed the “odd”)
Eight billion eighteen million eighteen thousand eight hundred eighty five.
(Leave out the eight hundreds, except the last).
[nevermind]…
You got it. Or at least that’s the answer I had. It’s a tricky little puzzle and it’s a while since I looked at it.
:smack: Bravo!
Eight billion eighteen million eighteen point nine with a bar over it
?
Nice–and sorry for doubting you! 
(Begbert, I don’t hardly think that follows the puzzle’s nomenclature for integers…)
Details, details. 
In my defense for not solving this one, I saw the first step and said, “forget it! That’s WAY too much work for solving a puzzle.” I’m pretty damned impressed with those of you who did it.
Here a couple of easier favorites that have not been mentioned yet:
++++ What’s so special about the number 143?
If you take any random three digit number (example: 847) and you repeat the pattern to create a six digit number (Ex. 847847), the resulting number is always evenly divisible by 143. Always. So what’s the deal with 143?
++++ Another Car Talk favorite:
A guy working at the corner Exxon notices a customer who stops in each morning and buys exactly 50 cents worth of gas. Sometimes (but not always) the same customer even comes back later the same day…again purchasing only and exactly 50 cents worth of gas.
The employee can’t help but shake his head at the guy’s behavior. Finally, after watching this customer over and over, the Exxon employee approaches the chap and says, “I see you come here at least once every day. May I suggest a simple repair that would prevent you from having to come here so often. Indeed, You might not have to come back at all.”
What’s the repair?
This one’s easy. 143 divides evenly into 001001, so of course it divides evenly into any multiple of it. 143 just happens to be the largest proper factor of 1001, so it looks more impressive than simply saying “The resulting number is always evenly divisible by 001001”.
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You have 1000 100g weights and 1000 99g weights, each indistinguishable from each other. You also have a balance scale, which has a left and right pan and only tells the difference in weights between the two pans. How can you guarantee finding two different piles of weights, each of the same size, such that the weight of each pile is different using only one weighing of the balance scale?
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You have a backyard with a circular jacuzzi and rectangular pool. How can you draw one straight line through your backyard that cuts the area of the jacuzzi and the pool in half?
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You have a circular pepperoni pizza with 2n slices. Each slice has some natural number of pepperoni slices. You and your friend eat the pizza, one slice at a time. You choose a slice and take it. Then your friend takes one of the two open slices, to the left and the right of the hole. Then you take one of the two open slices, and so forth. How can you guarantee getting at least half of the pepperoni slices on the pizza?
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A mouse nibbles through 9 111 blocks of cheese, stacked like a 333 rubiks cube. It starts at the top left of the cube. Can it eat every block of cheese and end up in the center of the cube?
For problem 4, wouldn’t a 3 * 3 * 3 cube contain 27 blocks? Perhaps I misunderstand the setup.
Supposing problem 4 was meant to involve 27 blocks with the mouse starting at one corner of the cube and only moving from blocks to orthogonally adjacent blocks, No, you cannot; imagine the blocks colored alternatingly black and white; the mouse must eat blocks in alternating color order, and thus, in eating 27 blocks, must start and stop at the same color. However, the corners of the cube have the opposite color from the center
For problem 2, Use the line passing through the centers of the circle and rectangle
otorophile’s puzzles:
2:Find the center of the pool and the center of the jacuzzi, and draw a line connecting those points. This seems too easy, though, so I suspect I’m missing something
3:Again, this seems too easy… Start by taking the piece with the most pepperoni, and then on every choice take the piece which has more. Is there some reason that won’t work?
4:OK, on this one, I see the trick. No, it can’t, because once it’s eaten all of them, there’s no cube left to be in the center of.
Chronos, your #3 solution doesn’t guarantee you’ll get at least half. For example, consider pizza like the following: 3 0 2 0 2, arranged in a circle. I start by taking 3, the opponent takes the available 2, I have to take a 0, the opponent takes the other 2, I lose.
One actual solution is this:
Color the slices alternatingly black and white. Whichever color has at least half the pepperoni, take one from that color on each turn. The opponent is forced to take one from the other color each turn; thus, you get all of your chosen color, guaranteeing you at least half
For problem #1,
Stick 500 weights on each side. If they don’t balance, you’re done; otherwise, collect them all into one pile, and put the previously unused weights in the other pile. (This works because if they balance, their total weight is 100kg - an even number of grams, which means that among the unused pile of weights, the total is 99kg + an even number of grams; the difference between these two is 1000 - a multiple of 4 grams, which can’t be 0 as 1000 isn’t divisible by 4)
Er, wait, clearly that last bit in the last post is embarrassingly wrong…
Haha, isn’t 3^3 9? You are right though.
No, that’s it.
Don’t be greedy.
Nope, pretend the mouse is in space.
Think about the most information you can get from one weighing.
One more: You have 2n apples and 2n bananas arranged in a circle, in some arbitrary order. Can you always draw a line through the circle such that on each side of the line there are n apples and n bananas?
bras–>brass, cares–>caress. Both of which seem to imply their own jokes/wordplay but the exact phrasing eludes me.
For problem #1,
Set one weight aside; divide the rest into three groups of 666. Weigh two of these groups: if unequal, you are done. If equal, you know the remaining group is either heavier or lighter.