OK....tried to use the spoiler and it didn't work.

Sorry folks....I know the answer, too, so if you don't want to read it, scroll quickly past the Asteriks*****

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If she's "about to be" 61, that means she's still 60. Obviously that's how come you can say she's "double" your age. You aren't speaking litterally, of course - You're just saying that 60 is double 30, so that's the "double" you're talking about.

If you recently turned 30, and she's just about to turn 61, that means when you turn 31, she'll be just about to turn 62. That will mean she WILL BE 62 at least for some period of time when you ARE 31. That will be the only other time she will ever be "double" your age. From then on out, you just creep closer and closer to being her age!

***OK, I didn't see that you actually added dates to this. Of course this answer all has to do with IF you had asked this about 4 days ago BEFORE she actually turned 61. (Now she IS 61.)

Before she turned 61, she was 60 and you were 30. Double (although not EXACTLY double.) Now she turned 61 and you're still 30. Not double. Soon you'll be 31 and she'll still be 61. Still not double.

But NEXT YEAR, AFTER October 6th, she will be 62 for about a month while YOU are still just 31. That number is once again double.

Again - It's not EXACTLY double. She will only be EXACTLY double ONE TIME in your entire lifetime. That has already past. (I'm too tired to figure out the exact date.)

Saperd aubot 2 ihecns wtroh of snad all oevr the folor, or jsut mkae him filp the cion on the bceah. It's bonud to lnad on its sdie a few tmeis!

I don't believe the second one is right. (I have no idea about the first.)

It would depend on what is possible for a minimum payment. If you have to use conventional methods, and the minimum is 1 of ANY form of real currency, then of course the answer is infinity. If, however, you can go to fractions of the smallest currency, then the number approaches $3.00.

OK - forget what I said. I was thinking of something different. I was thinking 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32......in which case the amount would approach $2.00, not $3.00.

For this problem, it needs to be 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8..... right?

I'm not quite sure how to solve this one, but I'm sure it'll be easy for someone.

lim(a->inf) (1/(a*log2(1+0.1/a))) ~ 6.93147 years

since harmonic series is a divergent one, X will end up with infinite amount of money

I don't believe the second one is right. (I have no idea about the first.)

It would depend on what is possible for a minimum payment. If you have to use conventional methods, and the minimum is 1 of ANY form of real currency, then of course the answer is infinity. If, however, you can go to fractions of the smallest currency, then the number approaches $3.00.

I'm sure there is an easier way to say this, but here is what I think the formula basically is:

2/2 + 4/4 + 6/8 + 8/16 + 10/32 + 12/64 + 14/128 + 16/256 + 18/512 + 20/1,024 + 22/2,048 + 24/4,096 + 26/8,192 + 28/16,384 + 30/32,768.......etc., etc., etc......

It appears to me that the number approaches 4.

The reason I think thoughtfulfellow is probably onto the right idea is because I know bonanova is always deep into the mathematical intricacies of problems.

I'll go way out on a limb and try this from a "trick question" point of view nonetheless.

I'll say Davey was the man who paid the most money since Alex is a woman whose whole name is Alexandra.

One, or even zero if the mouse doesn't live longer than a year (considering the 13days maturity time). Except if the first little girl mouse finds a way to clone herself!

Remember he said, "A man goes to te store and buys a girl mouse." Provided he fed this GIRL mouse and tended to it well, there is only going to be ONE mouse! (Zero if the mouse died.)

I believe this has been posted before.

A hint about "Hints" - Hints are not meant to give the entire answer. This was a fairly easy one to figure out. Let folks figure it out. Then if they can't figure out the answer, post a "Hint", or post the actual answer.

I won't bother leaving the simple answer since you already left it.

There are exactly 2,652 two-card combinations. 52 * 51

Correction – it would be 2,652 combinations only if the ORDER of the two cards mattered. Divide that by 2, and that’s the number of combinations without consideration of the order of the 2 cards.

There are exactly 1,326 two-card combinations without regards to the order of the cards. (52 * 51) / 2

How many of those two-card combinations contain exactly 1 ace?

Ace of clubs plus any of 48 non-Ace cards.

Ace of diamonds plus any of 48 non-Ace cards.

Ace of hearts plus any of 48 non-Ace cards.

Ace of spades plus any of 48 non-Ace cards.

That’s 192 two-card combinations that include exactly one Ace.

Deal 8 cards to 4 players. It doesn’t matter how. Let’s take each player 1 at a time.

Player A has a 192 out of 1,326 chance that he has one of those two-card combinations.

Assume that he does. Now there are only 50 cards left in the deck, only 3 of which are Aces. There are only 1,225 two-card combinations left [ (50 * 49) / 2 ], and only 3 Aces times only 47 non-Ace cards to create the right hand. That’s 141 two-card combinations left that contain exactly 1 Ace.

Player B, therefore, has a 141 out of 1,225 chance that he has one of the remaining two-card combinations of a single Ace.

Assume that he does. Now there are only 48 cards left in the deck, only 2 of which are Aces. There are only 1,128 two-card combinations left, [ (48 * 47) / 2 ], and only 2 Aces times only 46 non-Ace cards to create the right hand. That’s 92 two-card combinations left that contain exactly 1 Ace.

Player C, therefore, has a 92 out of 1,128 chance that he has one of the remaining two-card combinations of a single Ace.

Assume that he does. Now there are only 46 cards left in the deck, only 1 of which is the remaining Ace. There are only 1,035 two-card combinations left, [ (46 * 45) / 2 ], and only 1 Ace times only 45 non-Ace cards to create the right hand. That’s 45 two-card combinations left that contain exactly 1 Ace.

Player D, therefore, has a 45 out of 1,035 chance that he has one of the remaining two-card combinations of a single Ace.

Multiply all the chances together since they all MUST happen to meet your criteria.

192/1,326 * 141/1,225 * 92/1,128 * 45/1,035 = Your Answer

It has a 0.0059100563302243974512882075907286% chance of happening.

Put another way, it will happen only once per every 16,920.3125 deals.

The first formula ends up giving you a 1 out of 8,484.4375 chance. I'm just not sure where I'm going wrong.

I THOUGHT I was right until I read the first person to answer. If I were a betting man, I'd probably figure I was most likely to be wrong. At any rate, there MUST be an easier way to do this problem than the way my wee brain drew it up. Can someone enlighten me as to where (if at all) I went wrong, and what the easier way to think of this problem is?

Thanks all! I enjoy this forum even though it's just a bit over my head.

6 = VI

40 = XL

64= LXIV

Of course! Good job!

1. A third of six is the "X" in "S-I-X". "X" is 10 in Roman numerals.

2. Nothing yet...

3. Nothing yet...

4. If you cut the figure "8" in half horizontally, it's a "0". Multiply that by anything to get zero.

3. Take the letters is "S-i-x-t-y" and subract those letters from the word "S-e-v-e-n-t-y-s-i-x" and you're left with "S-e-v-e-n". Seven is an odd number.

I tried to use similar logic for #2, but 40 is spelled "Forty", so I can't combine "Forty" with "six" to make "Sixty-four". Doesn't work.

There is a succession of passengers, beginning with Paul, whose choice of seat is not guided by a boarding pass. Either the pass is misplaced, as in Paul's case, or someone [that would be Paul, or the one in whose seat Paul sat, or ... etc.] has already taken the passenger's seat. The other passengers all take their own seats.

In the exceptional cases, however, there are exactly two events that will seal the final result right then and there. Moreover, these are the only events that affect the outcome.

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1. The passenger chooses Paul's seat.

2. The passenger chooses Peter's seat.

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In the first case, Peter will get his own seat. In the second case, he will not.

The only thing to understand, therefore, is that these two events have exactly the same probability, namely p=1/Q, where Q is the number of empty seats at the time the choice is made. They are unlikely choices at the start, where the number of seats to choose from is large, and they become more likely as the plane fills up. But they are the only choices that matter, and they are, always, equally likely.

The chance that Peter gets his own seat is thus inescapably 1/2.

Thank you VERY much! This explanation was a very clear way to illustrate the concept. I appreciate the extra time you gave here! Thanks again!

All that matters is this:

Does someone sit in Paul's seat before Peter boards?

If so, Peter gets his seat. Otherwise he get's Paul's seat.

Those are the only places Peter can sit.

The answer is independent of N.

You can work it out for small values of N - it's always 1/2.

For very small values, like N = 2 and N = 3 I can fully understand why it would be 1/2. As soon as N becomes higher than that, wouldn't the probability start falling?

Just imagine N = 4. The first man does NOT sit in his seat because he lost his pass. (He had a 1/4 chance of NOT sitting in his seat in the first place.) Now Passenger #2 has a 1/3 chance it was his seat that was just taken, and if it WAS taken, he has a 1/3 chance of sitting in the final passenger's seat, a 1/3 chance of sitting in the first man's seat, and a 1/3 chance that he'll sit in Passenger #3's seat.

Then Passenger #3 comes on. Now IF either the first man OR Passenger #2 took his seat, he has a 1/2 chance of taking the final passenger's seat.

I'm sure I'm missing something in my logic. It seems to me that the more passenger's there are on the plane, the lower the probability that the last man's seat would be available.

Can you help me understand more clearly why I'm wrong?

the judge.

The postman and the plumber both are wearing nametags as part of their uniform, and neither of the nametags says "David".

Assume a deterministic strategy that wins in N cases out of 32. Then after 32 rounds with all the 32 possibilities you get N*1000 + (32-N)*(-50) = 1050*N-32*50 = 1050*N-1600.

Now if N>=2 then the strategy is positive in the long run.

Simplest positive strategy with N=2 is to always guess the color of the person to the left of you. This always wins in 2 out of 32 cases (RRRRR and BBBBB). Therefore it wins 500 dollars per 32 games, in the long run.

Found a rather asymmetrical strategy with N=4. Not sure if N>5 strategies are possible (I was not able to get past N=4).

I am not great with these puzzles, but your post got me thinking of something:

If you ONLY guess the person to your left, then that does give you 2 out of the 32 whenever everyone is Blue or everyone is Red. But that means, on the times when the person to your left is a DIFFERENT color than the person to your right, you have a ZERO chance of winning. Even if YOU are right about your color being the same as the person to your left, you know that the person to your right will end up being wrong by writing down your color which is similar to the color of the person to your left.

What would happen if you always planned on writting down the same color as the person next to you ONLY when BOTH your neighbors had the same color as you?

When your left neighbor is wearing an opposite color as your right, you ought to still give yourself a little bit of a chance, so you should guess randomly that time.

I just can't mathematically figure out how much more of an advantage that gives you if you plan on guessing during the times when your neighbors are wearing opposite.

Tuesday.

Jagan is lying by saying he didn't lie on a Monday, (yesterday), but that's OK because he's supposed to lie on Tuesday's.

Pradeep is telling the truth about not lying on a Monday, (yesterday), and he is supposed to be telling the truth on Tuesdays.

The relative size of the triangle and sphere determines the angles.

More precisely, the solid angle subtended by the triangle is what matters.

On a sphere the size of the earth or the size of a marble, either one,

a triangle with vertices all on a great circle has a 540^o sum,

for both the internal and external angles.

Ah....I'm fairly sure my wee brain gets it now!

Thanks much for the further explaination, Bonanova!

Ok, here's the question simplified. Would it be a triangle?

Again - going completely by what Bonanova said, "YES". This would be a Spherical Triangle as defined in Spherical Geometry. In Spherical Geometry, a Spherical Triangle is a "normal" triangle.

If you posted this must follow the definitions in Euclidean Geometry, then of course the answer is "no".

See Bonanova's post for the links.

Right, can you draw a 180* triangle on a sphere the size of a tennis ball?

Bonanova said the triangle on a sphere would be between 180 and 540 degrees. I THINK this means that the larger the sphere, the closer the triangle's angles would add up to 180 degrees; the smaller the sphere, the closer the triangle's angles would add up to 540 degrees.

One simple solution

We draw a straight line starting from the magnetic north pole to as far as we can go for more accuracy (ignoring the Earth's curvature). Now the compass is pointing north along the line we just drew. at a right angle from the line, we head west or east drawing a perpendicular line on our original line. We keep on walking and drawing until the compass has moved a certain degree. We then have a golden L shape which is half our desired rectangle so the rest will simply be completing the rectangle.

The key here is to calculate this specific angle at where we stop. Actually we have two choices and one is more accurate than the other. The better choice is making the second line the longer one. To calculate the angle, we use simple trigonometry. Theta= tan-1(Phi) where Phi is the golden ratio=1.618....

I hope I was clear without any drawings

This puzzle is talking about a GEOMETRIC COMPASS. Not a compass that finds North/South/East/West.

One of these: http://www.dailyclipart.net/wp-content/uploads/medium/Math1.jpg

(Great puzzle. Well beyond the compass of my ability!)

if N = infinity.... distance travelled = sqrt(2)

else distance travelled = 2

To be clear, compared to you all I have the mathematical capabilities of a squirrel. With that in mine, here is my question regarding Abhijith's post:

It can't just go from being 2 to being the sqrt (2) when it reaches infinity, can it?

Shouldn't it start approaching sqrt (2) when it starts approaching infinity?

Hmm....then again, at what point in time could it "start" approaching infinity?

The more I think about it, the more I think I'd have to agree with Abhijith's post as is. As long as infinity is never actually reached, which of course it can't be, then it's 2. If you assume infinity, then it's the sqrt (2) like he said.

OK...I'll just look forward to reading the other answers like I usually do.

Good question, mmiguel1!

cube 1 - 0,1,2,3,4,5

cube 2 - 0,1,2,6,7,8

The 6 serves as a 9 upside down

It IS necessary to have a "0" on both dice since you want to be able to display ever low number as 01, 02, 03...08, 09. Just a note about the question, however, you will never actually use "00".

The only point at which the bridge could break with the additional weight of a bird is the very middle of the bridge where it is most vulnerable. The birds landed on the truck about 3/4 of the way through the trip, so there was no danger.

That coupled with the amount of fuel lost by the time the truck is more than 1/2 way. (Taken from the first answerer.)

You'll just die or hunger and thirst? Probably better to let the blades hit you....

OK, I'll point out the obvious. You just replied to someone about how they should use Spoilers, then you didn't use a spoiler for your answer!

I liked the idea of moving to the middle and perhaps even laying down. My concern would be that there is always one part of the blender piece that actually touches the bottom of the glass. Isn't that part right in the middle???

I think perhaps laying down near the sides of the glass would give you the best chance to survive. The only worry there would be that the air would move so fast that you'd be swept up by the breeze.

Someone posted that this was retarded. I agree with folks that said a post like that is mean and shouldn't be part of this thread.

On the other hand, I can see where this person may have been coming from.

This question is the first post from a brand new person. I would think that someone would have at least put a little more thought into their question. I'm not intending to sound mean, but this is truly the sort of riddle that you'd find on the back of a box of Frosted Flakes.

At least follow the sorts of questions that are usually asked here for a while. The questions asked are usually quite challenging and the asker generally puts in a good deal of effort, and has at least some sort of original thought.