2 VERY indredibly hard mathematical problems.

[Dreamlax]

So many viewers not enought answers.... tsk tsk.

[Starman]

Q1. 0

[Yonatan]

1) Zero, because of the multiplication by (x - x).
2) South Pole?
Or, perhaps the distant planet's surface area (or whatever it's called) was less than 10x10 miles so the North Pole isn't the only place where you can go 10s 10e 10n and get back to where you are?

It's probably something completely and totally different, but I'll get over it.

[Starman]

Q2. They are somewhere on the equator of a rather small planet, radius about 7.385 miles.
Driving South 10m they reach a latitude where the horizontal circumference is exactly 10 miles. Drive 10m East or West from here and they return to the same spot. Driving North again they reach the base.
The search from the North pole for a 10 mile radius does not extend to the equator so the base is not found.

Or the planet has no magnetic pole and the buggy's compass points to some metal at the base camp.

[Starman]

A correction to my earlier post:

The planet can be large, the base camp is in the southern hemisphere, 10 miles north of the latitude where they can circle the south pole in exactly 10 miles.
This easterly part of the journey actually takes them nowhere and the last 10 miles north return them to the camp.

[Guv]

There are an unbounded number of places near the south pole where you can go 10 miles south, ten miles east (or west) and ten miles north, ending up at your starting point.

For example, Consider a place near the South Pole where a circle of constant latitude is 5 miles in circumference. Any spot ten miles north of that lattitude would fit the conditions.

Then consider places near the South Pole where a circle of constant latitude has a 10 mile, 2.5 mile, or 1.25 mile circumference.

On earth, if they said they saw a few bears while taking their trip, it would have to be the North Pole because there are no bears near the South Pole.

[Dreamlax]

Damn you maths geniuses. Trust you to spoil it for anyone who can't work it out........ me. I'm slow.

But there isn't an unbound number of places. Eventually you'd get to numbers so small you couldn't drive the difference, if you know what I mean. So Mr. Maths, I have one which will have you stumped (well maybe not. Try this:

Fill in the missing number:

10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31, 100, ?????, 10000

And yes, it is possible, and it only has ONE answer (unless you are a super Mr. Maths who can find out another.)

[Starman]

Quote: So many viewers not enought answers.... tsk tsk.

Sorry, I thought you were chasing for an answer.


Can you let us know which professor thought what?


As to your missing number I'm stumped - perhaps you could provide a hint in a few days time ?

[Yonatan]

Originally posted by Dreamlax
Fill in the missing number:

10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31, 100, ?????, 10000

Well if you give each number an index, where the first number gets the index 16 and the last one gets the index 2, you get:

10 (16), 11 (15), 12 (14), 13 (13), 14 (12), 15 (11), 16 (10), 17 (9), 20 (8), 22 (7), 24 (6), 31 (5), 100 (4), ????? (3), 10000 (2)

All these numbers are actually 16 in a different number system, whose base is in the parentheses.
So, ????? is 16 in base 3.

Answer: 121

[Dreamlax]

Stop cheating Yonatan. You would've never have guessed to start indexing at 16 and finish at 2. If you know these, don't spoil it for the people who don't.

The first professor said that if you were 10m north of where you could drive 10m right around the planet, and get back to the same place. The second professor said that he could be wrong, because you could drive around the planet twice (5m circumference).

The reason why it couldn't be a planet in our solar system is because we only have one sun. We would've realised if we had driven around the planet because it would turn dark (except Neptune (or the one which has an almost horizontal axis) I guess).

[Dreamlax]

One last one, see if you can replace the letters with numbers:

      S E N D
    + M O R E
---------------
    M O N E Y

And if you know it, don't make it look like you know it.

[Active]

ok.

mth.plaekim/tg7js6s/moc.loa.sresu : lru siht ot og t'noD

[Starman]

If more than one letter can represent the same number you could have:

9000
1000
10000

[Guv]



   SEND
 + MORE
------------
  MONEY

The following is a start on a solution, assuming decimal arithmetic.

M must be one (cannot carry more than one when adding two numbers).

S must be 8 or 9 (S + 1 caused carry).

Looks like O should be zero.

[Active]

The rules Dreamlax did'nt mention :

Rules which every alphametic obeys:

1. The mapping of letters to numbers is one-to-one. That is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.

2. The digit zero is not allowed to appear as the left-most digit in any of the addends or the sum.

There is only one and unique soluion.
If you lost hope solving this ...read my last post.

[Active]

Ok...here is another (Solved) Example


    SMART
+MEMBER
 -------
  ACTIVE

has the solution


   93457 A=4 B=6 C=1 E=2 I=0 M=3 R=5 S=9 T=7 V=8
+323625
-------
 417082

[Dreamlax]

I don't even know the solution to the puzzle myself!

[Active]

Drum rolls please....

Introducing the best Alphametic Puzzle Solver on the Internet.

To add to the suspense....the url is in a txt file.{So that we can get some statistics...hehe }

[Starman]

That's a pretty clever solver Active.

It say's there are 28 possible solutions to send+more=money in base 16.

My own rushed effort took about an hour to find the single (legal) base 10 solution - but you did get to see it thinking.
I like to make my computer work hard for it's living and if it takes longer to run the program than I took to write it then I feel that the computer is working for me rather than the other way round.