And now: ANSWERS!
Jun. 30th, 2015 12:31 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
theweaselkingThe answers to Yesterday's Quiz Post! If you haven't already looked, go do that now, I'll wait.
PUZZLE #1:
"For any prime number greater than 3 (labelled "P") why is it that P2-1 is ALWAYS evenly divisible by 24?"
This one has two catches to it. The first, and biggest, is that you need to remember from grade-school algebra that (P2-1) = (P-1)(P+1)
Which is to say, P2-1 is the same as "the product of the numbers one lower and one higher than P"
Catch #2: when we say something is evenly divisible by 24, what that means is that the number contains 24 as a factor, which means it contains 2*2*2*3 as prime factors.
So, then, what do we know about the numbers P-1, P, and P+1?
We know P is prime, which means it is odd, which means P-1 and P+1 are both EVEN numbers - each one *must* have 2 as a factor.
We now P-1 and P+1 are two consecutive even numbers, which means one of them *must* have 4 (or 2*2) as a factor.
Finally, we know that P-1, P, and P+1 are three consecutive integers, which means that one of them *must* be evenly divisible by 3.... and it's not P. Because P is prime, it can't be divisible by three. So either P-1 or P+1 is.
Putting this all together: We know that p2-1 is the product of P-1 and P+1. We can prove that both P-1 and P+1 have two as a factor, and that one of them must have a second 2, and that one of them (maybe the same one, maybe the other one) must have 3 as a factor. This means that their product *must* contain 2*2*2*3 as a factor, which is 24.
It's a clever bit of numberplay
PUZZLE #2
A vending machine dispenses two kinds of chocolate bar. It has three buttons - "Snickers", "Mars", and "Random". Snickers gets you a Snickers bar, Mars gets you a Mars bar, Random gets you either a Snickers or a Mars, 50/50 chance. At least, that's how it's SUPPOSED to work, but the guy who set it up screwed up the labels so none of the labels on the buttons are right. It costs $1 for a chocolate bar; what's the minimum amount of money required to be able to know, for certain, which button is which?
You only need to spend $1 to correctly and certainly know what all three buttons do.
The catch here is that *none* of the labels are right, and you know this. So your solution is to toss in $1 and press "Random".
You know the Random button *can't* be random, so it must be either Mars or Snickers. Whatever you get, that's the button.
Assume you get a Mars. So your buttons are now:
Random -> Mars
Mars -> ?
Snickers -> ?
Your Mars and Snickers buttons give either Snickers or Random.... and you KNOW none of the buttons are right, which means the Snickers button *does not* give Snickers. So the Snickers button must be Random, and that leaves the Mars button with only one option.
The same thing happens, of course, if your first press gets you a Snickers. "Random" is Snickers, "Mars" can't be Mars, therefore Mars must be Random and Snickers must be Mars.
PUZZLE #3 The old marble puzzle! You have 12 marbles, identical except one of them is either lighter or heavier than the others. You don't know which is the odd one out or if the odd one is light or heavy. You have only an old-style balance scale to use so you can measure marbles or groups of marbles against each other but not against anything else. Using only this scale, determine which is the odd one, and if it's light or heavy, in no more than 3 weighings.
Nobody got this one, and I don't blame you at all. It's one of the ones that's REALLY HARD to re-derive the trick, even if you know it. But it's still a clever puzzle.
This one's got to be done in steps.
Number your marbles 1 through 12, so you know which is which. if you think writing a number on the marble would be "cheating", then sort them in a row and just be careful to always put the right marble back in the right order, you pedant.
WEIGHING #1 is always, always going to be 1,2,3,4 versus 5,6,7,8. And what you do in response to that depends on if it balances.
If WEIGHING #1 BALANCES: The odd marble out is one of 9,10,11,12.
WEIGHING #2A is 1,2,3 (known-good marbles) against 9,10,11.
If WEIGHING #2A BALANCES: The odd marble out is #12. Weigh it versus any other marble and you will learn if it's lighter or heavier, and you're done in three measurements.
If WEIGHING #2A DOES NOT BALANCE: The odd marble out is 9, 10, or 11, and you can tell by which way the imbalance went if the weird marble is light or heavy Weigh 9 vs 10: If they balance, 11 is the odd one and you already know if it's light or heavy, you're done in three. If they don't balance, the one that's light (or heavy, you learned which already) is the outlier and you're done in three.
That's the easy part: If the outlier isn't part of weighing #1, you can get it in 3 weighs, no major tricks.
If WEIGHING #1 DOES NOT BALANCE: It's a little harder. You still don't know if the weird marble is lighter or heavier and you don't know which side of the scale it's on.
I'm going to assume 5,6,7,8 is heavier for this but the same process works in reverse if 1,2,3,4 is heavier.
WEIGHING #2B is 1,5,6 versus 2,7,8 : one marble from the "light" side and two from the "heavy" side, on each side.
If WEIGHING #2B BALANCES: The odd marble out is either 3 or 4, and it's light. Weigh 3 against any other marble, like 9: if they balance, 4 is the odd one and it's light. If they don't balance, 3 will be lighter than 9 and again, you're done in three measurements.
If WEIGHING #2B DOES NOT BALANCE: Each side has one "maybe light" and two "maybe heavy" marbles. So you know, if 2,7,8 is heavy, that EITHER 7 or 8 is a weird heavy marble, OR 1 is a weird light marble. So your third measure is 7 vs 8. If they match, 1 is light. If they don't match, the heavy one is it, and you're once again done in three.
(The same goes in reverse if 1,5,6 is heavy: Either 5 or 6 is weird and heavy, or 2 is weird and light, so weigh 5 vs 6.)
And you're done, in no more than three weighings.
PUZZLE #1:
"For any prime number greater than 3 (labelled "P") why is it that P2-1 is ALWAYS evenly divisible by 24?"
This one has two catches to it. The first, and biggest, is that you need to remember from grade-school algebra that (P2-1) = (P-1)(P+1)
Which is to say, P2-1 is the same as "the product of the numbers one lower and one higher than P"
Catch #2: when we say something is evenly divisible by 24, what that means is that the number contains 24 as a factor, which means it contains 2*2*2*3 as prime factors.
So, then, what do we know about the numbers P-1, P, and P+1?
We know P is prime, which means it is odd, which means P-1 and P+1 are both EVEN numbers - each one *must* have 2 as a factor.
We now P-1 and P+1 are two consecutive even numbers, which means one of them *must* have 4 (or 2*2) as a factor.
Finally, we know that P-1, P, and P+1 are three consecutive integers, which means that one of them *must* be evenly divisible by 3.... and it's not P. Because P is prime, it can't be divisible by three. So either P-1 or P+1 is.
Putting this all together: We know that p2-1 is the product of P-1 and P+1. We can prove that both P-1 and P+1 have two as a factor, and that one of them must have a second 2, and that one of them (maybe the same one, maybe the other one) must have 3 as a factor. This means that their product *must* contain 2*2*2*3 as a factor, which is 24.
It's a clever bit of numberplay
PUZZLE #2
A vending machine dispenses two kinds of chocolate bar. It has three buttons - "Snickers", "Mars", and "Random". Snickers gets you a Snickers bar, Mars gets you a Mars bar, Random gets you either a Snickers or a Mars, 50/50 chance. At least, that's how it's SUPPOSED to work, but the guy who set it up screwed up the labels so none of the labels on the buttons are right. It costs $1 for a chocolate bar; what's the minimum amount of money required to be able to know, for certain, which button is which?
You only need to spend $1 to correctly and certainly know what all three buttons do.
The catch here is that *none* of the labels are right, and you know this. So your solution is to toss in $1 and press "Random".
You know the Random button *can't* be random, so it must be either Mars or Snickers. Whatever you get, that's the button.
Assume you get a Mars. So your buttons are now:
Random -> Mars
Mars -> ?
Snickers -> ?
Your Mars and Snickers buttons give either Snickers or Random.... and you KNOW none of the buttons are right, which means the Snickers button *does not* give Snickers. So the Snickers button must be Random, and that leaves the Mars button with only one option.
The same thing happens, of course, if your first press gets you a Snickers. "Random" is Snickers, "Mars" can't be Mars, therefore Mars must be Random and Snickers must be Mars.
PUZZLE #3 The old marble puzzle! You have 12 marbles, identical except one of them is either lighter or heavier than the others. You don't know which is the odd one out or if the odd one is light or heavy. You have only an old-style balance scale to use so you can measure marbles or groups of marbles against each other but not against anything else. Using only this scale, determine which is the odd one, and if it's light or heavy, in no more than 3 weighings.
Nobody got this one, and I don't blame you at all. It's one of the ones that's REALLY HARD to re-derive the trick, even if you know it. But it's still a clever puzzle.
This one's got to be done in steps.
Number your marbles 1 through 12, so you know which is which. if you think writing a number on the marble would be "cheating", then sort them in a row and just be careful to always put the right marble back in the right order, you pedant.
WEIGHING #1 is always, always going to be 1,2,3,4 versus 5,6,7,8. And what you do in response to that depends on if it balances.
If WEIGHING #1 BALANCES: The odd marble out is one of 9,10,11,12.
WEIGHING #2A is 1,2,3 (known-good marbles) against 9,10,11.
If WEIGHING #2A BALANCES: The odd marble out is #12. Weigh it versus any other marble and you will learn if it's lighter or heavier, and you're done in three measurements.
If WEIGHING #2A DOES NOT BALANCE: The odd marble out is 9, 10, or 11, and you can tell by which way the imbalance went if the weird marble is light or heavy Weigh 9 vs 10: If they balance, 11 is the odd one and you already know if it's light or heavy, you're done in three. If they don't balance, the one that's light (or heavy, you learned which already) is the outlier and you're done in three.
That's the easy part: If the outlier isn't part of weighing #1, you can get it in 3 weighs, no major tricks.
If WEIGHING #1 DOES NOT BALANCE: It's a little harder. You still don't know if the weird marble is lighter or heavier and you don't know which side of the scale it's on.
I'm going to assume 5,6,7,8 is heavier for this but the same process works in reverse if 1,2,3,4 is heavier.
WEIGHING #2B is 1,5,6 versus 2,7,8 : one marble from the "light" side and two from the "heavy" side, on each side.
If WEIGHING #2B BALANCES: The odd marble out is either 3 or 4, and it's light. Weigh 3 against any other marble, like 9: if they balance, 4 is the odd one and it's light. If they don't balance, 3 will be lighter than 9 and again, you're done in three measurements.
If WEIGHING #2B DOES NOT BALANCE: Each side has one "maybe light" and two "maybe heavy" marbles. So you know, if 2,7,8 is heavy, that EITHER 7 or 8 is a weird heavy marble, OR 1 is a weird light marble. So your third measure is 7 vs 8. If they match, 1 is light. If they don't match, the heavy one is it, and you're once again done in three.
(The same goes in reverse if 1,5,6 is heavy: Either 5 or 6 is weird and heavy, or 2 is weird and light, so weigh 5 vs 6.)
And you're done, in no more than three weighings.