Math Problems

[Mercutio]

A small trick:

Let a=b

Then ab=b^2

Then ab-a^2=b^2-a^2

Factoring, we get a(b-a)=(b+a)(b-a)

Dividing out this (b-a), we get a=b+a

But b=a

Hence a=2a, or 1=2

What/where did it go wrong?


And also: No dog has five legs, but any dog has hour legs more than no dog; hence a dog has nine legs.

[Janny]

You divided by zero somewhere. I've already asked that one.
Yeah if a = b then a - b = 0.

[Beren3000]

Neat trick, Mercutio!
And your answer to my problem was correct, too!
He wouldn't lie and tell you the way to some other village because we just assume that villages A and B are the only ones around

[mithrand1r]

Form an expression that evaluates to 24 that contains two 3's, two 7's. You may use zero or more of the operators +, -, *, and /, and parentheses.

[mithrand1r]

Form an expression that evaluates to 24 that contains two 4's and two 7's. You may use zero or more of the operators +, -, *, and /, and parentheses.

[mithrand1r]

Form an expression that evaluates to 24 that contains three 5's and one 1. You may use zero or more of the operators +, -, *, and /, and parentheses.

[Janny]

Hello maths buddies. I'm wondering if you could help...

As a curious overlap between organic chemistry and discrete maths we were working on networks made from carbon (nodes order 4) and hydrogen (nodes order 1), in the ratio Cn Hn+2, ie. the alkanes.
Could anyone point me towards the formula which determines the number of isomers, or different networks, for each case of n?

Incidentally, is it true that what is chemically different could be mathematically similar in terms of networks? Our teacher (not a chemist) insisted that (sorry..) 2,3-dimethylpentane was mathematically similar to 2,4-dimethylpentane.

[Jonathan]

The ratio should be Cn H2n+2 for alkanes

We know that for the first three alkanes, there are no isomeres. Methan, ethan and propane can only have one appearance each. The 4th alkane can have two forms and the 5th can have three.

The number of isomers for the alkanes follows a certain sequence, called A000602 in the The On-Line Encyclopedia of Integer Sequences (OEIS). Its formula is a bit tricky to write but its first ten numbers (n=1,2,3...10) are
1, 1, 1, 2, 3, 5, 9, 18, 35, 75.
You can do a search on the sequence number A000602 at OEIS.

I don't know in which way 2,3-dimethylpentane and 2,4-dimethylpentane would be similar mathematically. They're isomers of n-heptane and according to the sequence above, they are two of 9 alkane isomers with 7 carbons each.

[Janny]

*Checks* Oh dear...

Yes, 2n+2... Thanks, I had found the OEIS but I didn't really follow it. It was in notation I didn't know and I didn't get the references to other sequences. Also, I'm working from feeble 16-year-olds' maths...

I only wanted the formula as a chemistry lesson curiosity... rekindled by the realisation by the teacher that the 62.5 trillion combinations later down our alkane isomer list weren't achieved through manual reconfiguration of molecule models...

As regards to the 2,3 and 2,4 differences, I think she thought that both had carbons attatched to the same number of carbons. ie. 4 carbons bonded to only 1 other, 2 that were attatched to three other Cs and 1 that was only attatched to 2. I think she didn't count the hydrogens, which meant all C nodes were order 4.

Anyway, thank-you.

[Radagast The Brown]

I have a little riddle to you...

Let's see... we have 100 prisoners. They're locked each in different cell (room? Chamber?). Everyday one of the prisoners is randomly chosen (A prisoner can be chosen millions of times when another was not chosen even once). He's led to another room - the room is empty except one lamp. He can either turn on/off the light, or keep it as it was. They have track of time.
The prisoners need to say when they're sure all of them have already been in the room at least one time. They can make a plan before they're locked. If they're wrong they die, if they're right they're freed.
Your job is to make the plan - so that at least one of them will know that all the others have been in the room - and their only contact after they make the plan in the beginning is the lamp in the room.

Do you understand? *hopefully*

[Jonathan]

I think I have a solution to the problem.

The prisoners have to agree that apart from one guy, all 99 of them except have to turn the lamp on only twice. That is, the first and second time a prisoner finds the lamp turned off, he turns it on. If the prisoner returns to the room with the lamp multiple times after that, he should just ignore the lamp.

Now one of the 100 prisoners has to turn off the lamp every time he finds it on and ignore it if it's already off. He has to count the number of times he turns the lamp off and on count 199, he can be sure that all the other 99 prisoners have visited the room.

If the prisoners were told that the lamp was initially turned off, they could just agree to turn the lamp on once and let one of them turn off the lamp every time until he counts 99.

This seems right, right?

[Radagast The Brown]

They do know in the beginning if the lamp is on/off, but it doesn't really matter to the solution.

You're right, yes. Do you know how much time other (like me) had to think, only to get to a 1 million years solution? Oh well. Go you. I must say it's impressive.

Let's see... another one, but a really easy this time. (the first one suppoed to be very hard. )
A driver is practising for a race.. He wants to do two laps in an average speed of 60 kph.(mph, it doesn't matter) He does the first one in the speed of 30 kph. What should be his speed to get to the 60 kph average?

[Elanor's Angel]

um 90



Any hoo... does anyone know the next 5 or so digits of PI? So far I have:
3.1415926535897932384626433832795028841971 Does anyone?

[Elanor's Angel]

Oh! Never mind I found it, It's 6939937510582 (next is 097, I think.)

[Radagast The Brown]

It's not 90. (it's the first thing everyone says)

[Jonathan]

That was a tricky problem indeed, at first I couldn't comprehend how the heck you could say it's not 90

The thing is - if you complete the first lap driving 30 kph and the second lap doing 90 kph, you will finish the second lat a lot faster than the first lap, ie. 3 times as fast. So timewise, you have spent 3/4 of the time on the first lap and only 1/4 on the second lap. That will give you an average speed of:

30 kph · 3/4 + 90 kph · 1/4 = 45 kph, not 60.

In order to get an average of 60 kph, you'd have to drive infinitely fast. If you can be bored with a mathematical explanation, here it comes
You can calculate the average speed for the whole race with the following formula:

_______Total distance (two laps)_______
Time for the 1st lap + Time for the 2nd lap

Let's assume that one lap is the distance D (a positive number). Then the whole race is 2D and the time it takes to complete the 1st lap is D/30. If the average speed of the 2nd lap is X, then the time for the lap is D/X. Use the formula above and you'll get the equation:

____2D___ =
D/30 + D/X

____2____ = 60 kph
1/30 + 1/X

Now you can see that the only way you can get an average speed of 60 kph is if the term 1/X is infinitely small, ie if X is infinitely big. The higher the speed during the second lap, the closer the driver will come to 60 kph. He will approach 60 kph asymptotically but he'll never reach it.

I hope someone follows.

[Janny]

Marvellous! Loverly maths.

And Elanor's Angel, Dr Kanada of the University of Tokyo calculated it to 1.2411 trillion digits ,here's the first ten thousand:
3.
1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989 3809525720 1065485863 2788659361 5338182796 8230301952 0353018529 6899577362 2599413891 2497217752 8347913151 5574857242 4541506959 5082953311 6861727855 8890750983 8175463746 4939319255 0604009277 0167113900 9848824012 8583616035 6370766010 4710181942 9555961989 4676783744 9448255379

[Janny]

*Deep breath*
7747268471 0404753464 6208046684 2590694912 9331367702 8989152104 7521620569 6602405803 8150193511 2533824300 3558764024 7496473263 9141992726 0426992279 6782354781 6360093417 2164121992 4586315030 2861829745 5570674983 8505494588 5869269956 9092721079 7509302955 3211653449 8720275596 0236480665 4991198818 3479775356 6369807426 5425278625 5181841757 4672890977 7727938000 8164706001 6145249192 1732172147 7235014144 1973568548 1613611573 5255213347 5741849468 4385233239 0739414333 4547762416 8625189835 6948556209 9219222184 2725502542 5688767179 0494601653 4668049886 2723279178 6085784383 8279679766 8145410095 3883786360 9506800642 2512520511 7392984896 0841284886 2694560424 1965285022 2106611863 0674427862 2039194945 0471237137 8696095636 4371917287 4677646575 7396241389 0865832645 9958133904 7802759009 9465764078 9512694683 9835259570 9825822620 5224894077 2671947826 8482601476 9909026401 3639443745 5305068203 4962524517 4939965143 1429809190 6592509372 2169646151 5709858387 4105978859 5977297549 8930161753 9284681382 6868386894 2774155991 8559252459 5395943104 9972524680 8459872736 4469584865 3836736222 6260991246 0805124388 4390451244 1365497627 8079771569 1435997700 1296160894 4169486855 5848406353 4220722258 2848864815 8456028506 0168427394 5226746767 8895252138 5225499546 6672782398 6456596116 3548862305 7745649803 5593634568 1743241125 1507606947 9451096596 0940252288 7971089314 5669136867 2287489405 6010150330 8617928680 9208747609 1782493858 9009714909 6759852613 6554978189 3129784821 6829989487 2265880485 7564014270 4775551323 7964145152 3746234364 5428584447 9526586782 1051141354 7357395231 1342716610 2135969536 2314429524 8493718711 0145765403 5902799344 0374200731 0578539062 1983874478 0847848968 3321445713 8687519435 0643021845 3191048481 0053706146 8067491927 8191197939 9520614196 6342875444 0643745123 7181921799 9839101591 9561814675 1426912397 4894090718 6494231961
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[Janny]

*Breathes again*
9699009019 2116971737 2784768472 6860849003 3770242429 1651300500 5168323364 3503895170 2989392233 4517220138 1280696501 1784408745 1960121228 5993716231 3017114448 4640903890 6449544400 6198690754 8516026327 5052983491 8740786680 8818338510 2283345085 0486082503 9302133219 7155184306 3545500766 8282949304 1377655279 3975175461 3953984683 3936383047 4611996653 8581538420 5685338621 8672523340 2830871123 2827892125 0771262946 3229563989 8989358211 6745627010 2183564622 0134967151 8819097303 8119800497 3407239610 3685406643 1939509790 1906996395 5245300545 0580685501 9567302292 1913933918 5680344903 9820595510 0226353536 1920419947 4553859381 0234395544 9597783779 0237421617 2711172364 3435439478 2218185286 2408514006 6604433258 8856986705 4315470696 5747458550 3323233421 0730154594 0516553790 6866273337 9958511562 5784322988 2737231989 8757141595 7811196358 3300594087 3068121602 8764962867 4460477464 9159950549 7374256269 0104903778 1986835938 1465741268 0492564879 8556145372 3478673303 9046883834 3634655379 4986419270 5638729317 4872332083 7601123029 9113679386 2708943879 9362016295 1541337142 4892830722 0126901475 4668476535 7616477379 4675200490 7571555278 1965362132 3926406160 1363581559 0742202020 3187277605 2772190055 6148425551 8792530343 5139844253 2234157623 3610642506 3904975008 6562710953 5919465897 5141310348 2276930624 7435363256 9160781547 8181152843 6679570611 0861533150 4452127473 9245449454 2368288606 1340841486 3776700961 2071512491 4043027253 8607648236 3414334623 5189757664 5216413767 9690314950 1910857598 4423919862 9164219399 4907236234 6468441173 9403265918 4044378051 3338945257 4239950829 6591228508 5558215725 0310712570 1266830240 2929525220 1187267675 6220415420 5161841634 8475651699 9811614101 0029960783 8690929160 3028840026 9104140792 8862150784 2451670908 7000699282 1206604183 7180653556 7252532567 5328612910 4248776182 5829765157 9598470356 2226293486 0034158722 9805349896 5022629174 8788202734 2092222453 3985626476 6914905562 8425039127 5771028402 7998066365 8254889264 8802545661 0172967026 6407655904 2909945681 5065265305 3718294127 0336931378 5178609040 7086671149 6558343434 7693385781 7113864558 7367812301 4587687126 6034891390 9562009939 3610310291 6161528813 8437909904 2317473363 9480457593 1493140529 7634757481 1935670911 0137751721 0080315590 2485309066 9203767192 2033229094 3346768514 2214477379 3937517034 4366199104 0337511173 5471918550 4644902636 5512816228 8244625759 1633303910 7225383742 1821408835 0865739177 1509682887 4782656995 9957449066 1758344137 5223970968 3408005355 9849175417 3818839994 4697486762 6551658276 5848358845 3142775687 9002909517 0283529716 3445621296 4043523117 6006651012 4120065975 5851276178 5838292041 9748442360 8007193045 7618932349 2292796501 9875187212 7267507981 2554709589 0455635792 1221033346 6974992356 3025494780 2490114195 2123828153 0911407907 3860251522 7429958180 7247162591 6685451333 1239480494 7079119153 2673430282 4418604142 6363954800 0448002670 4962482017 9289647669 7583183271 3142517029 6923488962 7668440323 2609275249 6035799646 9256504936 8183609003 2380929345
9588970695 3653494060 3402166544 3755890045 6328822505 4525564056 4482465151 8754711962 1844396582 5337543885 6909411303 1509526179 3780029741 2076651479 3942590298 9695946995 5657612186 5619673378 6236256125 2163208628 6922210327 4889218654 3648022967 8070576561 5144632046 9279068212 0738837781 4233562823 6089632080 6822246801 2248261177 1858963814 0918390367 3672220888 3215137556 0037279839 4004152970 0287830766 7094447456 0134556417 2543709069 7939612257 1429894671 5435784687 8861444581 2314593571 9849225284 7160504922 1242470141 2147805734 5510500801 9086996033 0276347870 8108175450 1193071412 2339086639 3833952942 5786905076 4310063835 1983438934 1596131854 3475464955 6978103829 3097164651 4384070070 7360411237 3599843452 2516105070 2705623526 6012764848 3084076118 3013052793
2054274628 6540360367 4532865105 7065874882 2569815793 6789766974 2205750596 8344086973 5020141020 6723585020 0724522563 2651341055 9240190274 2162484391 4035998953 5394590944 0704691209 1409387001 2645600162 3742880210 9276457931 0657922955 2498872758 4610126483 6999892256 9596881592 0560010165 5256375678

Naturally, I've personally verified it.

[Radagast The Brown]

Yes, you're right Jonathan. It's a tricky one. But it's logically - if you do the first lap in 1/2 of the average time, then you do it two times slower... meaning you have no time to do the second lap.