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[color=purple]One, Two, Three, Four, Five, ......Hundred are a hundred words which do no contain the letters A, B, C, and D. Can you think of ten more?
Am I being immeseley stupid here? Doesn't the word "hundred" contain 2 x "d"? (Though it doesnt contain A,B,C AND D). As for thinking of any more words that don't contain A, B, C and D, I could give you a few thousand!
I guess I must have missed the joke (or fallen for it)!
I always thought that andtidisestablishmentarianism was the longest real word.
Whatever next? Maybe someone on this forum can show the probability "proof" that life exists elsewhere. Alas, there is none. All that exists is us and our incurable desire not to be alone.
***5
Ok, here goes, maybe a little clumsy and probably not 100% proof, but here goes:
for values of a,b,c greater than or equal to 1:-
If a, b & c all = 1, then (a + 1)^7 * (b+1)^7 * (c+1)^7 = 2,097,152 and 7^7* a^4 * b^4 * c^4 = 823,543
any increase in a,b or c can be viewed simply as a comparison between the functions (a+1)^7 and a^4. For all positive numbers greater than 1, this only increases the divergence. Therefore, it is only necessary to consider values of a,b,c between 0 & 1.
As a,b,c tend to zero, [(a+1)^7....] tends to an minimum value of 1, while [7^7.....] tends to zero. Again, as a,b,c increase above zero, the effect on the comparison is the same as for values above one. Namely, any increase in (a+1)^7 will be greater than the corresponding increase in a^4.
Obviously, negative values would reverse this, but for all positive values, the expression must be true, though I cannot actually see a point at which the expressions could ever be equal!
ganesh, I am seriously struggling with the proof of ****3. I can work through the logical structure of the 4 given equations and I can see an obvious similarity with the expression to be solved. However, for the life of me, I just cannot get an expression that absolutely links this expression to those funtions of e. The problem is that the numerator's progression is a funtion of (1+2n)^2 rather than x^n, as given in the examples, which is destined to diverge greatly as n increases.
My solution was simply based around common sense, in that the answer HAD to have a relationship to e, otherwise the examples are pointless!. Doing a sum of the first few values of the expression soon led to an answer of 14.5914, which is clearly 1 + 5e. Equally, the progression follows a very similar pattern to the formula for e, so it was clear that it is not necessary to bring in high values of x as their effect is negligable.
The worked example that gives this missing link would be greatly appreciated!
That's me on the bottom row - the fat cat (or whatever it actually is). A bit worried now about suggesting its a cat cos we all know that espeon isn't a pink bunny after all, so maybe that isn't a cat either!
Sorry, jU, but giving further explanation now would mean turning this forum into a biology lesson!
OK, scrub the plumbline - the moments don't work like that for an irregular shape! My mistake. Soooorryy!
fgarb, my reasoning for choosing the 5/6 solution was based on the way ganesh had drawn the diagram. Note, the vertical lines are conveniently punctuated with dots at 1/6 intervals. Therefore, a straight line joining the dot below J with the dot below E would do the job just fine!
2. Grammar check, please! I presume that the question should be "exactly 2 EQUAL parts" (ie, 2.5 square units each).
If so, I have 2 answers, both a little lateral and probably not what you are looking for:
a) a horizontal line 5/6 units up from the bottom giving 5/6 of 3 squares below the line = 2.5 units
b) I would employ physics.... cut the shape from a homogenous material, suspend it from any point along the shape's edge, and run a plumbline to find the centre of gravity line! As the material is homogeneous, the area each side of that line would be 2.5 square units. I would guess that there would be many solutions where this COG line would pass entirely within the shape to give a valid solution.
My solution to 1 was slightly different from JLFM, but the same principle. It is, of course, a perfect example of thinking "outside the box"!!
Questions 2...... hmmm, need some thought....
What a lovely compliment! I wish that my teacher had agreed with you in 1978 (or, even the examination board!!). I was far from the perfect student and my maths was only average at best. But what I have done in life since then is to learn from mistakes, work from first principles wherever possible and always try to understand WHY things are the way they are. It may take a lot longer than remembering formulae, but it is much more sound in the long-term.
Alas, question ***3 is currently beyond me and I really need to dust away a lot more cobwebs before I can suggest a properly worked solution, but I do agree with 1 + 5e being the correct answer, based on observation rather than on algebraic proof.
Thanks, Krassi.
You were obviously posting your answer while I was still working on mine, so I saw yours after posting mine! I had got as far as multiplying out and simplifying, but I just hadn't got as far as the final factorisation. It used to be so easy when I was 18, as I was doing it all the time, but you don't have much need for this kind of maths as a hotelier, so it takes me a bit longer to get my thinking in gear!!
Krassi, my work now... I own a hotel & restaurant. But for most of my working life I have been an accountant & general businessman/entrepreneur
Wait until September. All will be revealed!!!
***2
Well, the answer is , but my algebra is so rusty that I am struggling to come up with the elegant solution rather than the trial and error version! Give me time and I WILL get there!
Tigeree, as I am over 13 (!), I feel quite comfortable telling you I am from the UK. Originally an Essex boy, raised in Surrey and now living in Devon with a wife and 1.3 children. Hope I don't break any rules by telling you this!!
Try this:
[((3n)^2 x 5/9) -3n] / [3n x (3n-1)]
Mathsy, I agree with your logic, but the question asked for two numbers to be selected at random, not the same number to be selected twice. Hence, I excluded the possibility of x=y. This has the effect of removing 3n from both numerator and denominator. Progressing through the series, up to very high values of n will still tend towards 55.55%, but always slightly less than this and, at very low values of n, significantly less.
A technicality, maybe, but if I am playing poker, I like to know how many aces are in the deck!
Michael Shumacher, of course!
This sounded like a trick question at first, but having spent a little time on it, the pattern is somewhat surprising and gives a remarkably high number of results divisible by 3!
The answer I get seems to tend towards , but it does vary depending on the value of n. (eg, if n=1, then it is 33.33%, as I had expected).
Having worked on this the hard way, let's now try to get a formula.
Krassi: I think we have found the limit of my Russian! I also don't speak it too well and Bulgarian would definitely cause me a problem (though I know it is similar).So let's not go any further! By the way, yes, I also love Russia and its people - it is a beautiful country that has suffered much, but its people have the same capacity for warmth and love as anywhere else in the word. It's cultutral and intellectual history are nothing short of awesome.
Why, thank you, espeon! But any idea who's helmet it is?