Re: Maths problems (interesting ones)
 

23) What number comes next: 77, 49, 36, 18...

7x7=49
4x9=36
3x6=18
1x8=8 ;)

Re: Maths problems (interesting ones)
 

Here's an interesting one I picked up a while back - the solution is pretty neat:

You have been commissioned by the town to place four colored lamp posts in a circular area of a park. The lamps are red, green, blue, and yellow. You must place them inside of the circular area, which will later be fenced in. The town tells you to place the lamps in an orientation so that a person walking all the way around the circular fence looking inward can see every possible arrangement of colors ordered from left to right in their field of view.

Is there any orientation of lamp posts that makes this possible? If not, what is the maximum number of arrangements the person would be able to see? Prove it any way you can.

Top view of circular area attached:

Re: Maths problems (interesting ones)
 

26)

You can only get 14 combinations.

With 4 lights there are 24 combinations:
RGBY RGYB RBGY RBYG RYGB RYBG GRBY GRYB GBRY GBYR GYRB GYBR BRGY BRYG BGRY BGYR BYRG BYGR YRGB YRBG YGRB YGBR YBRG YBGR

The problem can be simplified as such: Draw four points on a plane and then draw a line to connect every two points. Then draw a circle enclosing the four points. The lines will divide the outside of the circle into a number of regions. Since each region represents a different order of lights, the goal is to get the maximum number of regions outside of the circle.

When you put 4 points that do not make any parellel lines or have 3 points on a single line, you get 18 regions. 4 of them are inside of the points, so you can have up to 14 outside the circle. 2 of them will start far outside of the circle, so only 12 will be on the perimeter. It is impossible to get all 24.

Here is a new one that is neat:

27)

A camel works for a banana plantation that is 1,000 miles away from a market. This year, the harvest contains 3,000 bananas that need to be moved to the market. The camel can carry 1,000 bananas at a time, however, the camel eats one banana for every mile it walks. What is the maximum number of bananas from the harvest that the camel deliver to the market?

Re: Maths problems (interesting ones)
 

Bingo. I didn't even think about the two outside the circle...that's interesting.

Re: Maths problems (interesting ones)
 

Problem 21: 1000!

You are right, 249 trailing zeroes. Here is the concret value, calculated by a self written little java program.

1000! = 40238726007709377354370243392300398571937486421071 46325437999104299385123986290205920442084869694048 00479988610197196058631666872994808558901323829669 94459099742450408707375991882362772718873251977950 59509952761208749754624970436014182780946464962910 56393887437886487337119181045825783647849977012476 63288983595573543251318532395846307555740911426241 74743493475534286465766116677973966688202912073791 43853719588249808126867838374559731746136085379534 52422158659320192809087829730843139284440328123155 86110369768013573042161687476096758713483120254785 89320767169132448426236131412508780208000261683151 02734182797770478463586817016436502415369139828126 48102130927612448963599287051149649754199093422215 66832572080821333186116811553615836546984046708975 60290095053761647584772842188967964624494516076535 34081989013854424879849599533191017233555566021394 50399736280750137837615307127761926849034352625200 01588853514733161170210396817592151090778801939317 81141945452572238655414610628921879602238389714760 88506276862967146674697562911234082439208160153780 88989396451826324367161676217916890977991190375403 12746222899880051954444142820121873617459926429565 81746628302955570299024324153181617210465832036786 90611726015878352075151628422554026517048330422614 39742869330616908979684825901254583271682264580665 26769958652682272807075781391858178889652208164348 34482599326604336766017699961283186078838615027946 59551311565520360939881806121385586003014356945272 24206344631797460594682573103790084024432438465657 24501440282188525247093519062092902313649327349756 55139587205596542287497740114133469627154228458623 77387538230483865688976461927383814900140767310446 64025989949022222176590433990188601856652648506179 97023561938970178600408118897299183110211712298459 01641921068884387121855646124960798722908519296819 37238864261483965738229112312502418664935314397013 74285319266498753372189406942814341185201580141233 44828015051399694290153483077644569099073152433278 28826986460278986432113908350621709500259738986355 42771967428222487575867657523442202075736305694988 25087968928162753848863396909959826280956121450994 87170124451646126037902930912088908694202851064018 21543994571568059418727489980942547421735824010636 77404595741785160829230135358081840096996372524230 56085590370062427124341690900415369010593398383577 79394109700277534720000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 000000000000000000

The upper calculation cost 2,5 seconds. And the calculation for 10000!, which has 2499 trailing zeroes, runs about 2,5 hours on a normal PC.

Best regards
Carsten

Re: Maths problems (interesting ones)
 

Prove that theta is a good approximation for sin(theta) when theta is small. Prove that 1+x is a good approximation for e^x when x is really small. They are both the same problem and involve some advance calculus. The first problem is really important since it drastically simplifies the pendulum calculations if theta is really small.

Re: Maths problems (interesting ones)
 

Starting with the second problem:

Maclaurin series e^x to get
e^x = 1 + x + x^2 / 2! + ... + x^n / n!

Take the limit of that series as x->0 and you'll notice that it equals 1, aka e^0 = 1.

Supposing that x is not zero, but infinitely close to zero, you have 1 + x + (small number) / 2 + (smaller numbers) / (bigger numbers).
The smaller number over a number approaching infinity is zero.
Therefore, you have 1+x for small values of x.

Similarly.... sin (theta) = theta for small values of theta. Do a Maclaurin... and you'll see as x gets increasing small, all other terms of the series except x become 'nothing.'


Not the best proofs by any shot... but they explain the problems. :)

Re: Maths problems (interesting ones)
 

My own quote
That took me all of five seconds to figure out the answer. They are the same thing except the maclaurin series is a taylor series when a=0.

Re: Maths problems (interesting ones)
 

I think this thread can use a little revival.

The unsolved puzzles are:

25)
27)
Any new puzzles would be appreciated. Thank you.

Re: Maths problems (interesting ones)
 

27) 500 bananas

28) Here's one I thought of at breakfast on Sunday, after seeing a watch upside down:
The minute hand of an analog clock is pointing directly at 12, while the hour hand is halfway between 1 and 2. This is not a valid state for the clock to be in, probably because the clockworks have rotated relative to the face (the part with the numbers). How far do I have to rotate the face of the clock, and in which direction, to make the time valid, and what time is it, to the nearest second?

For bonus points, name all the possible times that this arrangement of hands could be, if I rotate the face all the way around.

Also, please note that I'm not an experienced creator of math problems, so the answer isn't pretty.

Re: Maths problems (interesting ones)
 

can you explain the answer to27 please?

Re: Maths problems (interesting ones)
 

The camel starts out with 1000 bananas. He walks 250 miles towards his goal, drops 500 bananas, and walks back. He takes 1000 more bananas from the stock.

Then he walks 250 miles toward his goal, picks up 250 more bananas, and keeps walking. Now he holds 1000, there are 250 bananas at 250 miles, and 1000 bananas at home. He walks 250 more miles, drops 500 bananas, and walks all the way back, using the 250 bananas he has plus the 250 more he has stashed at 250 miles. Then he takes the last 1000 from his original stash, and walks all the way to his goal, picking up 500 extra when he passes 500 miles.

Re: Maths problems (interesting ones)
 

ahh cunning!

Re: Maths problems (interesting ones)
 

That is close, but you can get even more bananas out of this method.

28)

I am fairly sure I have the right answers here. You can set up the equation that has T for the time in minutes and X for the difference between the hour hand and the minute hand in rotations as -11T/720 = X. This becomes -11T = {90, 810, 1530, 2250, 2970, 3690, 4410, 5130, 5850, 6570, 7290}, as you substitute 0.125, 1.125, 2.125, etc... for X (because 0.125 is the distance from 12:00 to 1:30, and as time goes on, you have to add full rotations). All the info is written here.