Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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I think you are correct, mathsyperson. Thank you. I do tend to make mistakes such as those.

**iheartmaths**- Replies: 2

Hi guys and gals,

I posed this problem to myself one night and quickly discovered that it is less innocuous than it seems! Graphically, the solutions are 2, 4, and a negative transcendental number (-0.7666646959...). However, out of curiosity I decided to see if there was an algebraic solution. From what Ive gathered on the internet, there does not seem to be one. Also, as can be easily seen, the problem has two obvious solutions (2, 4), of which I was only capable of finding x equals 2. Thus, I am convinced that I did something wrong.

I am sorry if my LaTeX is a little rough. I have just learned it through Dross' crash course. It was very confusing to interject text into the mathematics. So, I will explain any ambiguous steps if needed (at least until I get better), but somehow I doubt the need will arise.

Otherwise, would anyone care to comb it over to help me spot the mistake?

Thank you very much in advance.

Why did the chicken cross the Moebius strip?

To get to the other side... wait, no, that wasn't it....:)

Assassin's Creed rocked. May I suggest the Orange Box, merely for Portal.

**iheartmaths**- Replies: 1

So, I was in a coffee shop one day thinking to myself how much a ton of money would be, and I came up with this. Feel free to critique, review, advise, teach me something new. I'd appreciate it.

v = td(p/m)

where v is the value,

t is equal to a standard U.S. ton, or a short ton, I believe;

d is the denomination,

p is equal to a U.S. pound,

and finally m equal to the mass of a single bill of d denomination.

a word about m: obviously, once would need some sort of average, since the mass varies quite a lot. My guess is that after something like 10,000 bills, the average would be pretty stable. So, if anyone knows exactly what that number is, let me know.

So, for ease of calculation, I'll make m = 0.5, which is a gross overestimation.

v = 2000(1)(16/0.5)

v = 2000(32)

v = 64,000

Like I said, I just thought of this while bored, and thought it would be cool to play around with.

Daniel123 wrote:

Zach wrote:The second one's answer is: You're first. You've just lapped the guy in last.

The person in second etc can't lap the guy in last?

I think the important thing to notice is the wording. Now, obviously, this is only my opinion, but the word overtake implies to change position in respect to the race standings, not just to physically surpass someone. So, I would say that you're right. Besides the obvious confusion caused by the fact that every racer minus the person in last can indeed overtake the person in last by lapping them, and that subsequently, the person in last can overtake the person in second to last while still being in last, the wording and context of the question seem to suggest that they meant your race position (1st, 2nd, 3rd, etc.), in which case lapping would have done nothing.

Ricky wrote:

Probability only makes sense when there is a concept of chance. Pi's digits are determined whether you know them or not.

Determined? yes, I agree; known? Not at all. At least not after *n* number of decimal places. I'm not, however, saying that this is a valid counter-arguement, because I don't subscribe to the thinking of if we can't disprove it, it must exist. I will however still believe that in the higher decimal places if pi, its digits become a faith, in a sense. What I mean is that after *n* digits, we have to accept the fact that there is still an inifinite number of digits left. So, with that said, probability could very well be involved in pi, in that the division could take a turn for one million continuous zeros.

It's actually because the mind adds those values of 10 as values of 100. Look at this to illustrate the difference:

1000 + 400 (which the mind adds as 400 for some reason, even though the real value is 40. the same goes for all of the 10 values) + 1000 + 300 + 1000 + 200 + 1000 + 100 = 5000

1000 + 40 + 1000 + 30 + 1000 + 20 + 1000 + 10 = 4100

ganesh wrote:

Proving this is the most difficult, most frustrating to any mathematician.

Theorem :- The number pi does not ever contain a string of 1 million continuos zeros after the decimal.

Corollary 1 :- No irrational number ever contains a string of 1 million continuos zeros after the decimal.

Corollary 2 :- No irrational number ever contains a string of 1 million continuos any of the numbers from 0 to 9 after the decimal.

Attempt proving or disproving this, by counter-example or otherwise!

Now, I'm quite the amateur, so correct me when I veer off the mathematician's path here, as I most likely will do. If pi is non-repeating, non-terminating, then as many digits of pi that you can calculate is only equal to ∞-n/∞ of its digits, right? So, if we looked at pi and said, "Well, it cannot contain one million zeros. And look, we've calculated it one billion places, and no zero or string of numbers 0 through 9." Then aren't we haunted by the fact that this is only ∞-n/∞ of its digits, and the fact that there are infinite numbers left. So, even if we calculated it to some astronomically high digit place and found nothing, we'd have to calculate even more, and more, and more, forever. Well, interestingly enough, I would think that the probability of 14159265358979 is equal to 111..., or 222.... However, if we accept the fact that it is infinite, then however counter-intuitive it seems, doesn't the same probability go for a million zeroes? This is only my unprofessional opinion. But it would be interesting to discuss.

However, my friend, another armchair mathematician, brought up an interesting point. He said that if pi is a division problem, such as circumference divided by diameter, then in order for it to contain one million continuous zeros, pi would have to be terminating, or at least repeating, in that to contain one million zeros, circumference/diameter = 3.14 ...000..., which means it couldn't possibly jump back to 1, right?

Either way, I think it's another interesting point.

Welcome, relax4now.

Vietnam?

I hope you get a chance to discuss some of those unbelievable ideas.

Zdravstvuite, Umbros. I'm sure you'll like it here.

MasterofDisguise wrote:

iheartmaths wrote:I am afraid of

ALL insects, and I mean all of them. Also, heights, tight, enclosed spaces; In general, human beings. Where I live, it's not exactly the safest place in the country. So, once when I went out to walk and think, when I was much younger, I was robbed and my friend was beaten down with brass knuckles, for absolutely no reason might I add. So, ever since then I've been like an abused dog with humanity. I tend to stay away from all people and be a bit of a loner. I'm also afraid of snakes, long and dark hallways, the ocean, and dying early.Are you scared of flies??

im not scared of ANY insects(excluding beetles) unless bees are insects!

Oh, yes. They're all disgusting. And the noise the fly makes is terribly frightening to me. Have you ever thought of how sinister that sounds? That buzz? That drives me insane.

as in 25 x 11 =

take the first digit, which is 2 and write it down. Then add the first digit to the second digit, which is 2 + 5, which equals 7, so that now we have 2 & 7. Finally, take the last digit, which is 5, and write it down. So, 25 x 11 = 275

The good thing is that this method works for any number times 11, like 1,450,623. Just repeat the second step until you've done all the digits. So, 11 x 1,450,623 = 15,956,853

bossk171 wrote:

iheartmaths wrote:bossk171 wrote:we need someone who isn't a political robot.

I wonder if that's possible in this country.

I think that'd be Obama (just a little plug for my guy), but I also think that Kucinich and Richardson (who just dropped out) may also be free thinkers.

Definitely not Hillary.

Yeah, I've never been a fan of Hillary. I need to sit down one afternoon and do my research on the canidates since I haven't been paying much attention. Regardless, I know Hillary isn't the one.

hotgal24 wrote:

How to solve this Question?

:1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1

1/z means 1 over z a fraction.

The letters can be any number.

How to find seven numbers that fit into the equation?(the numerator is always 1)

What if its 8 sets of fraction? is there any technique?

Well, if you're allowed to use mathsyperson's technique, then 1/number of fractions that sum up to 1 would be your technique. So for 8 sets of fractions, it's just 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 = 1

Of course the reason this works is simply that if seven fractions sum up to 1, then the denominator is 1/7 of 1.

**iheartmaths**- Replies: 0

http://youtube.com/watch?v=nKq6_vjrxMo

Pi and e are on a blind date. Need I say more?

JohnnyReinB wrote:

Nevermind. Iheartmaths, I would appreciate it if you use hide tags:

{hide=inserttitlehere} blahblahblah{/hide}

use [ instead of {

The above would be

All right. Sorry, JohnnyReinB, I didn't know how to use the hidetags. Thanks.

So, i'll re-type my answers and delete my last post.

bossk171 wrote:

we need someone who isn't a political robot.

I wonder if that's possible in this country.