AB*C.DE=AB+C.DE. Assume that A through E are >=0, and further that A!=0. Then AB>=10.

Then C!=0, because otherwise AB*C.DE<AB+C.DE. Since C.DE<10 by assumption, if C>=2 then AB*C.DE>2*AB but AB+C.DE<2*AB so AB*C.DE<AB+C.DE. Therefore C=1 necessarily.

Now we have AB*1.DE=AB+1.DE. Rewrite this as AB(1+0.DE)=AB+AB*0.DE=AB+1.DE. => AB*0.DE=1.DE
=> AB=1DE/DE multiplying through by 100.

Now we need to find a fraction of the form 1DE/DE whose quotient is a two digit integer AB. Since AB>=10, this immediately restricts the available options for DE to anything <12. We observe that DE=02,04,05,10 all yield two-digit integers and therefore represent solutions to the original equation, but only DE=04 yields a solution for which all five digits are unique (the other fractions yield integers with a '1' which is also what C is).

Therefore AB=26=1.DE/0.DE=1.04/0.04 and this is necessarily the only solution in which all digits are different.

The 3 other solutions, incidentally, are
AB=51=1.02/0.02
AB=21=1.05/0.05
AB=11=1.10/0.10

QED, I think.

28, 32, 34, 32, 34, 36, 40, 38, 38, 40, 36, 38, 40

** **

1. The sequence could also be:

Starting with the 2nd and 11th prime (3 and 31), these numbers are the differences between successive pairs of primes...

The probability that you'll get 500,000 heads is very small but more likely than getting 499,999 heads and 500,001 tails, and so on. The exact odds of it are

Easy... sort of. The concept is simple but by picking a number as large as a million you create an awful lot of work to figure out the result.

The question is the same thing as asking, "Of all the million digit binary numbers, how many have exactly 500,000 1's?"

Let r = # of million digit binary numbers, so r= 2^1000000.

Now let n = the # of million digit binary numbers with half a million 1's. We can find n with the "choose function", you would say "one million choose half-a-million" and the formula for this is 1000000!/500000!*500000! That's a huge number, but a lot smaller than 2^1000000.

So the answer is n/r. Unfortunately those are some incredibly large numbers so normal tools are insufficient to compute them. Since I have a Macintosh I have access to the powerful Unix tool "bc" which can handle integers of virtually any length, but even for "bc" these kinds of calculations could take hours.

Before we have our computers spend all night figuring out 1,000,000 factorial, let's solve the problem with a much nicer value of r, let's say "1000 coin flips." My computer quickly digests these numbers and comes back with the probability .0252, or 2.5 percent.

Next let me try something more extreme, let's say 10,000 coin flips. After a much longer pause, it comes back with the answer .0079, or 0.8 percent. It's gone down!

Finally let's try 20,000 coin flips. What are the odds of getting exactly 10,000 heads? Now it take almost a minute to come up with the answer, and that answer is .0056, still lower.

<Ohio> <Per the motorcycle, I think the handlebars were carried off with the curtain. One of the models appears to be standing on a box which wasn't there before, so some portion of the bike is hidden in it.> Are you speculating that this is some kind of George Jetson motorcycle than can be taken apart almost instantly and parts of it collapse into a briefcase? No sir, you are way off. In fact let me say that the concept is very old--from the 19th century--it has nothing to do with any radical technology.