Math Is Fun Forum

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6^1953125 ends in: 02220063994332456747346268935437376264942880349499173996030029353486977990646384748721787109376

Well, I've hit the limit. I typed 2^195312500 into bc, and it merrily spit "Runtime error (func=(main), adr=15): exponent too large in raise" back at me. There must be an open-source calculator out there that could do it...I may just need to have a look.

Add six...pschaw. Add three!

10^6 = million
10^9 = billion
10^12 = trillion

Why, you say? Because it fits with the SI system's tendency to add 10^3 to its incremental units. And it's easier to say that a "Gigawatt" is a billion watts than it is to call it a thousand million (which doesn't quite make sense).

Or it could be because I'm American, and refuse to say "colour".

How did you know I'd be up all night writing a research paper? >:[

Maybe tonight...

Interesting. My first impression was, "What?" Then, "Oh. Cool!" Then, "There's a lot more times that you must just guess." Then, "I wonder if there are some mad techniques for guessing?"

For example, in regular minesweeper, I would take guesses based on a rough estimate of the probability of a mine (or blank) occupying a certain space. I based it on the number of bombs I knew were in a given area, the total number of bombs left, and total number of squares left. It worked quite well.

I haven't found any similar technique here. However, I do feel a lot less like a computer could do a better job at it than I could.

Bugs I've seen so far:
I found that I could drag a flag onto an already-flagged square to run down the number of available flags. I didn't test what happens when it hits zero; I'm guessing you get stuck in sweeper limbo.

Also, it's possible to click and explode mines that have been exposed by a shovel.

It also ran very slow, but that could just be because I'm running 2^39062500 in the background.

Mechanical suggestions:
Why limit the number of flags at all?
A "bombs remaining" counter would be nice.
Dragging flags is incredibly tedious. Couldn't you do a platform-neutral modifier-click, like shift, to flag mines?

Visual suggestions:
The horizontal lines are easy to follow, but the diagonals got a little hard on the eyes. Once you get a mishmash of flags out there, there's quite a lot of visual ambiguity in the diagonals--what I mean to say is, the eye has a hard time following the line, and you have to check closely and convince yourself that you're still on the same diagonal you started with. Alternating shading would be awesome.

It would be very useful to grey out or otherwise identify lines that have been "satisfied" so you can block them out of visual scans.

That's it for now.

Your first equation is correct. It's in the form y=mx + b, where m is the slope and b is the y-intercept. b was easy to find because it was given to you, but not all problems are so nice. In this case, point-slope form (y = y1 + m(x - x1)) is your friend. So:

y= 3 + 2(x - 2)
y= 3 - (x - 4)
y= 1 + 4.5(x - 2)

etc.

2^7812500 ends in: 719540313499900586420401787109376

Oh yeah, logarithms. Duh. Nice job!

Assuming that all servings contain an equal measure of all parts, you can average the calories accross the pieces.

Volume of a cylinder: V = πr²H
r = 1/2d = 9 3/8 = 75/8 in.
H = 2(3.5) + 5/8 = 7 5/8 in. = 61/8 in.

Total volume of the cake: π(75/8)²(61/8) = 2105.388631 in³

Volume of chocolate frosting: V = π(9)²(1/4) = 63.61725124 in³

Volume of buttercream frosting: We'll need three cylinders for this one.
V of top = π(9)²(3/8) = 95.42587685 in³
V of sides = [ π(75/8)²(61/8) ] - [π(9)²(61/8) ] = 165.0624687 in³
total V of b. frosting: 260.4883456 in³

Volume of cake portion = 2[ π(9)²(3.5) ] = 1781.283035 in³

Calories from cake: 1781.283035 * 140 = 249379.6248 cal
Calories from chocolate frosting: 63.61725124 * 320 = 20357.5204 cal
Calories from buttercream: 260.4883456 * 470 = 122429.5224
Total calories in cake: 392166.6676 cal

Number of 800-calorie servings = 392166.6676/800 = 490, with a little less than a fourth left over.

All that talk of cake is making me hungry...

2^12500 ends in: 38413079360370299109376
2^62500 ends in: 35077828320112175538998220106163281524347109376
2^312500 ends in: 2616427893377447294587109376
2^1562500 ends in: 28458279155722394465726629493874312276145787109376

That last answer was about 125 pages long! (Well, terminal screens anyway)
Here's a puzzler for you. If the program wraps the answer to 69 numbers per line, how many lines are in the answer to 2^1562500?

I wouldn't know how to solve that!

2^2520

39408424552214162695348543183638915172819172249751642655322154182349\
33676588009610655644786388200003560563883371670355420740089454019139\
50236214360506399705231203021164366069389563701733455174652493802096\
52827965938125948350891617678251689261632215488187059650565457777432\
98081872565023704682568753763162781359379857881608885188091378378731\
80086327183792757748702946460720720770436177477377229784500022657580\
65723362838393013791461968400922079126708976855218290361860314695008\
42192427800725780716480012657266798737517723023431143584285521349919\
38056446803917216962620267368806273089867659639177213488960155211698\
14921103068177978857814105435927428955641140043659870492782127521488\
14889702185765573255518895775073409289563384104009610960263526424138\
31783448576

Hey ganesh, what kind of computer are you using? Unix and lookalikes have a program called bc; it's an arbitrary precision calculator. Just for fun, I typed 2^312500 into it, and it spat out the answer, full and complete, in about 5 seconds. Not bad.

If you happen to be running Windows, there is a program called Cygwin (I think) that emulates a unixy environment. You might give it a look.