Best dating math problems ever

best dating math problems ever

Celebrate your partner this holiday season with the gift of math Behold: Cosmo's 65 Best Sex Tips Ever. Sex & Relationships. Why College Dating Is So Messed Up? 10 Reasons Why You Should Date a Nerd. The 8 Kinds of Guys You Lose Your Virginity To. 16 Really Girly Things Guys Absolutely Love to Do. The 19 Most Frustrating Things About Casual Dating. 25 Things Guys Have Really Got to Stop Doing, Right Now. nerdy guys.

best dating math problems ever

One of my favorites: Why is six afraid of seven? Because seven eight nine. And 9 looks a lot like 6 not standing on his head. The first sixteen digits of e, the base of natural logarithms, has a fascinating pattern: 2.718281828459045 in which “1828” is the year in which Andrew Jackson was elected president. What does that have to do with natural logs? Jackson was born in a natural log cabin. As for interesting problems, there are books with plenty of these.

Look into the probability of two people sharing the same birthday in a room full of randomly selected people, such as a college or high-school classroom. How many people would you need present for the chance to be 50%?

How many to be better than 90% One interesting (and also Deep )math problem is “squaring the circle”. Namely, build a square that has the same area of a given circle. People have tried for centuries to do This.

In the late 19th century, It was proven that Pi is a transcendental number ( i.e. Not the solution of a polynomial equation with integer coefficients) and hence the problem does not have a solution. This is interesting because of the negative answers. It is much harder to prove that something does not exist than coming up with a recipe ( proof) of this is how you do it. Take a triangle sides [math]p,q,r[/math] and angles [math]\alpha,\beta[/math] Then [math]p^{\alpha}q^{\beta}=r^{\alpha+\beta}[/math] is a meaningful condition.

The above condition is connecting a triangle with a complex solution of [math]\displaystyle a^z+b^z=c^z[/math] where [math]\displaystyle p=a^x,q=b^x,r=c^x[/math] Writing [math]z=x+iy[/math] we have actually an equivalent system [math]\displaystyle a^x\cos(y\log(a))+b^x\cos(y\log(q))-c^x\cos(y\log(c))=0[/math] [math]\displaystyle a^x\sin(y\log(a))+b^x\sin(y\log(b))-c^x\sin(y\log(c))=0[/math] And placing one by one expression to the right and squaring we get three equations that are revealing the law of cosines thus connecting a solution to a triangle: [math]\displaystyle c^{2x}=a^{2x}+b^{2x}-2a^xb^x\cos(\pi-y\log(\frac{a}{b}))[/math] [math]\displaystyle a^{2x}=c^{2x}+b^{2x}-2c^xb^x\cos(y\log(\frac{c}{b}))[/math] [math]\displaystyle b^{2x}=c^{2x}+a^{2x}-2c^xa^x\cos(y\log(\frac{c}{a}))[/math] Now if you observe [math]\cos(t)=\cos(-t)[/math] and simply take [math]\displaystyle y\log(\frac{b}{c})=\alpha[/math] [math]\displaystyle y\log(\frac{c}{a})=\beta[/math] then [math]\displaystyle \frac{\log(\frac{c}{a})}{\beta}=\frac{\log(\frac{b}{c})}{\alpha}[/math] and you have got as one of the possible connections (there are more): [math]\displaystyle a^{\alpha}b^{\beta}=c^{\alpha+\beta}[/math] To the [math]x[/math] and you have it [math]\displaystyle p^{\alpha}q^{\beta}=r^{\alpha+\beta}[/math] Not in a million years I could expect this funny connection in any context.

*Notice that angles are not necessarily between [math]0[/math] and [math]\pi[/math] meaning the angle could be in the form [math]h+2k\pi[/math]. I love the four 4’s. You can use any math-signs and you have to create the numbers from 0 to 100 using only four 4’s and all the symbols you know.

To make zero, you could write 4+4–4–4=0. For 1 you can use 4/4*4/4=1…. You can’t insert other numbers, only symbols. Extra: placing a point in front of a four is ,4=0,4. Using percentage:4%=0,04 and ,4% is 40 You can also write


best dating math problems ever

best dating math problems ever - Dating Advice for Women: The Deadbeat Boyfriend Test


best dating math problems ever

Committing to a partner is scary for all kinds of reasons. But one is that you never really know how the object of your current affections would compare to all the other people you might meet in the future. Settle down early, and you might forgo the chance of a more perfect match later on.

Wait too long to commit, and all the good ones might be gone. You don’t want to marry the first person you meet, but you also don’t want to wait too long. This can be a serious dilemma, especially for people with perfectionist tendencies. But it turns out that there is a pretty simple mathematical rule that tells you how long you ought to search, and when you should stop searching and settle down.

The math problem is known by a lot of names – “the secretary problem,” “the fussy suitor problem,” “the sultan’s dowry problem” and “the optimal stopping problem.” Its answer is attributed to a handful of mathematicians but was popularized in 1960, when math enthusiast Martin Gardner wrote about it in Scientific American.

In the scenario, you’re choosing from a set number of options. For example, let’s say there is a total of 11 potential mates who you could seriously date and settle down with in your lifetime. If you could only see them all together at the same time, you’d have no problem picking out the best. But this isn't how a lifetime of dating works, obviously. One problem is the suitors arrive in a random order, and you don’t know how your current suitor compares to those who will arrive in the future.

Is the current guy or girl a dud? Or is this really the best you can do? The other problem is that once you reject a suitor, you often can’t go back to them later. So how do you find the best one?

Basically, you have to gamble. And as with most casino games, there’s a strong element of chance, but you can also understand and improve your probability of "winning" the best partner.

It turns out there is a pretty striking solution to increase your odds. Wedding bands The magic figure turns out to be 37 percent. To have the highest chance of picking the very best suitor, you should date and reject the first 37 percent of your total group of lifetime suitors. (If you're into math, it’s actually 1/e, which comes out to 0.368, or 36.8 percent.) Then you follow a simple rule: You pick the next person who is better than anyone you’ve ever dated before.

To apply this to real life, you’d have to know how many suitors you could potentially have or want to have — which is impossible to know for sure. You'd also have to decide who qualifies as a potential suitor, and who is just a fling.

The answers to these questions aren't clear, so you just have to estimate. Here, let's assume you would have 11 serious suitors in the course of your life. 10/10 This method doesn’t have a 100 percent success rate, as mathematician Hannah Fry discusses in an entertaining 2014 TED talk. There’s the risk, for example, that the first person you date really is your perfect partner, as in the illustration below. If you follow the rule, you’ll reject that person anyway.

And as you continue to date other people, no one will ever measure up to your first love, and you’ll end up rejecting everyone, and end up alone with your cats.

(Of course, some people may find cats preferable to boyfriends or girlfriends anyway.) Another, probably more realistic, option is that you start your life with a string of really terrible boyfriends or girlfriends that give you super low expectations about the potential suitors out there, as in the illustration below. The next person you date is marginally better than the failures you dated in your past, and you end up marrying him. But he’s still kind of a dud, and doesn't measure up to the great people you could have met in the future.

So obviously there are ways this method can go wrong. But it still produces better results than any other formula you could follow, whether you’re considering 10 suitors or 100. Why does this work? It should be pretty obvious that you want to start seriously looking to choose a candidate somewhere in the middle of the group.

You want to date enough people to get a sense of your options, but you don't want to leave the choice too long and risk missing your ideal match. You need some kind of formula that balances the risk of stopping too soon against the risk of stopping too late.

The logic is easier to see if you walk through smaller examples. Let's say you would only have one suitor in your entire life.

If you choose that person, you win the game every time -- he or she is the best match that you could potentially have. If you increase the number to two suitors, there's now a 50:50 chance of picking the best suitor. Here, it doesn't matter whether you use our strategy and review one candidate before picking the other.

If you do, you have a 50 percent chance of selecting the best. If you don't use our strategy, your chance of selecting the best is still 50 percent. But as the number of suitors gets larger, you start to see how following the rule above really helps your chances. The diagram below compares your success rate for selecting randomly among three suitors.

Each suitor is in their own box and is ranked by their quality (1st is best, 3rd is worst). As you can see, following the strategy dramatically increases your chances of "winning" -- finding the best suitor of the bunch: The diagram compares your success rate for selecting randomly among three suitors.

Each suitor is in their own box and is ranked by their quality (1st is best, 3rd is worst) As mathematicians repeated the process above for bigger and bigger groups of "suitors," they noticed something interesting -- the optimal number of suitors that you should review and reject before starting to look for the best of the bunch converges more and more on a particular number. That number is 37 percent. The explanation for why this works gets into the mathematical weeds -- here's another great, plain-English explanation of the math -- but it has to do with the magic of the mathematical constant e, which is uniquely able to describe the probability of success in a statistical trial that has two outcomes, success or failure.

Long story short, the formula has been shown again and again to maximize your chances of picking the best one in an unknown series, whether you're assessing significant others, apartments, job candidates or bathroom stalls. Other variants of the problem There are a few tweaks to this problem, depending on your preferences, that will give you a slightly different result.

In the scenario above, the goal was to maximize your chances of getting the very best suitor of the bunch -- you "won" if you found the very best suitor, and you "lost" if you ended up with anyone else. But a more realistic scenario, as mathematician Matt Parker writes, is that "getting something that is slightly below the best option will leave you only slightly less happy." You could still be quite happy with the second- or third-best of the bunch, and you'd also have a lower chance of ending up alone.

If your goal is to just get someone who is good, rather than the absolute best of the bunch, the strategy changes a little. In this case, you review and reject the square root of n suitors, where n is the total number of suitors, before you decide to accept anyone. As in the formula above, this is the exact point where your odds of passing over your ideal match start to eclipse your odds of stopping too soon. For our group of 11 suitors, you'd date and reject the first 30 percent, compared with 37 percent in the model above.

All in all, this version means that you end up dating around a little less and selecting a partner a little sooner. But you have a higher chance of ending up with someone who is pretty good, and a lower chance of ending up alone. With a choice of 10 people, the method gets you someone who is 75 percent perfect, relative to all your options, according to Parker. With 100 people, the person will be about 90 percent perfect, which is better than most people can hope for.

9 scientifically verified ways to be more attractive In 1984, a Japanese mathematician named Minoru Sakaguchi developed another version of the problem that independent men and women might find more appealing. In Sakaguchi's model, the person wants to find their best match, but they prefer remaining single to ending up with anyone else. In this case, you wouldn't start looking to settle down until reviewing about 60.7 percent of candidates.

In this situation, you notice that, since you don't care too much if you end up alone, you're content to review far more candidates, gather more information, and have a greater chance of selecting the very best. These models are theoretical, but they do support some of the conventional wisdom about dating.

First, they offer a good rationale for dating around before deciding to get serious. Without a dating history, you really don't have enough knowledge about the dating pool to make an educated decision about who is the best. You might think your first or second love is truly your best love, but, statistically speaking, it's not probably not so.

Second, when you choose to settle down really depends on your preferences. If you want to find someone who is pretty good and minimize your chances of ending up alone, you'd try to settle down relatively early -- after reviewing and rejecting the first 30 percent of suitors you might have in your lifetime.

If your goal is to find the very best of the bunch, you would wait a little longer, reviewing and rejecting 37 percent of the total. And if you would like to find your perfect match, but you are also okay with ending up single, you'd wait much longer, reviewing and rejecting 60.7 percent of the total before you start looking for your match.

These equations are also reassuring for those with fear of missing out, those who worry about committing to a partner because they don't know what they might be missing in the future. The math shows that you really don't have to date all the fish in the sea to maximize your chances of finding the best. More about • You may not agree with our views, or other users’, but please respond to them respectfully • Swearing, personal abuse, racism, sexism, homophobia and other discriminatory or inciteful language is not acceptable • Do not impersonate other users or reveal private information about third parties • We reserve the right to delete inappropriate posts and ban offending users without notification You can find our Community Guidelines in full {{^nickname}} Community Guidelines • You may not agree with our views, or other users’, but please respond to them respectfully • Swearing, personal abuse, racism, sexism, homophobia and other discriminatory or inciteful language is not acceptable • Do not impersonate other users or reveal private information about third parties • We reserve the right to delete inappropriate posts and ban offending users without notification You can find our Community Guidelines in full {{^nickname}} About The Independent commenting Independent Minds Comments can be posted by members of our membership scheme, Independent Minds.

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best dating math problems ever

The answer is 2. Here is the reason: The number of 2 is the 21 number If you subtract the 7th number (2) by 6th number (0)= 2 If you subtract the 14th number (5) by 13th number (3) = 2 If you subtract the 21st number (x)by 20th number (0), it should give you 2 I don’t have a PhD yet. But I am passionate in Mathematics, and is interested in Game Theory in my Operational Research studies in my MSc.

It still took me 5 minutes in working it out. I could see why it is difficult. Simply because people would like to share the problem, but how about the solution? That is outside the square! You are welcome to visit my blog http://suifaijohnmak.wordpress.com I could make up these type of mathematical problems if you like. My email: suifaijohnmak@yahoo.com.au This is a method based on a difference in a series: i.e.

the difference between consecutive 6th and 7th and then 13th and 14th and then 20th and 21st numbers etc. At first glance, I couldn't find any relationship between the sum of the the numbers and the number sequence.

My deduction is then that there may be a series of "pattern" that repeat its sequence - i.e. either the sum or difference of 2 consecutive number in the "series", where there is no pattern in the first five, but repeated difference in the 6th and 7th number etc. This is similar to the nine dot experiement, where you need to join nine dots with 4 connected straight lines without leaving the paper, with the end of one line joining the start.

Have you tried that before? Concept is think beyond the square. I am not sure whether this is challenging or not, but I really appreciate you directing me to this. Hope you have every success in your PhD study. Great to meet you here. John I think David has got the correct explanation in Daniel's blog.

I have also included David's answer on my blog to avoid any confusion. Interesting to note that by comparing numbers could reveal something. I reckon this is not easily tackled by elementary school students. However, the logic behind could cause some uncomfortable feelings with children, especially if they don't see the "logic". This is something like computer logic based on a conversion system of a number to a binary digit system etc. I have left some comments in my blog as reflection.

Great to be "challenged" and re-think about learning after this experience. I may be the first to give reasons as I mentioned in my comment, though the reasons are not correct, in retrospect. It really makes me think of learning in the blogosphere, where the same situation may happen. When I first saw the problem, I did see the pattern of those 0 derived from a set of same numbers. I speculated that there was an operation or conversion in the set of numbers. However, I didn't think it matters much as it seems any equation could be operated like that.

May be that is often due to our mind set based on education - rationality and logic. It really doesn't work here. However, if kids have been trained to have such mind set once in their lesson, it is easy for them to transfer the learning to other similar problems.

So, one experience in "intuition" may help in unlocking some of the toughest math problem. I also noted that some of the IQ tests are not reliable measure, as one could improve the IQ score through repeated drills. I have tried the IQ test when I was young, and have since realised that one could improve upon repeated practice. I think IQ test could improve one's patterning skills. I think the "aha" moment coming out of the learning are equally important in establishing one's confidence in tackling similar life problems.

So, I think you collection of this problem is already a smart move. Dear Eugene, thanks for your comment. No, it is Russian. :-) Actually, in White Russia (or Belorussia), Russian is more widely spoken than Belorussian nowadays.

You can compare wiki articles about Belorussia, written in Belorussian http://be.wikipedia.org/wiki/Беларусь and in Russian http://ru.wikipedia.org/wiki/Белоруссия As you can see, in Russian there are no such letters as "i" or "ў" for instance. The former one is absent in the Ukranian as well... So, uh, I minored in math and can't make heads or tails of Sui's solution, and the Daniel's closed loops thing really throws me. But yeah, I came up with 2 too. I just noticed that if you let all the numbers on the left side of the equals sign be symbols instead of numbers (e.g.

saying that "0" is worth 1 since "0000" is equal to 4). Doing that you find that all primes are worth 0, "1" is worth 0, "8" is worth 2, and all other numbers are worth 1. So 2581 is 0 + 0 + 2 + 0 = 2. I mean, this sort of solution is consistent with all given left-hand sequences and it seems like something a child could do. Thanks for your comment, Cantay. This is true, one can "map" 0 to 1, 8 to 2, 9 to 1 and so on. This was proposed by David, in 11th comment to the Daniel's post: http://www.daniel-lemire.com/blog/archives/2009/01/10/finish-this-sequence-of-equalities/#comments Another issue is that you should forget about math, and consider numbers as unknown symbols.

Best, Misha Anonymous Hmm, I have a math/physics bachelors and a physics phd and I figured it out in under a minute. Your comment that preschool children could figure it out was a pretty big clue that it would be graphical rather than textual cues. It took a bit of mental squinting to stop reading things as numbers... guille roccuzzo its very simple the code has given in the examples, if you see the add of 1,2,3,4,5,7 is 0, so they don´t have value, but the add of 9 is 4 so its value is 1 each one of them, the 6 and the 0 is the same, and the 8 is 2, because of the first example given, which is 8809=6, the eights add 4, the 0 one, and one the nine, ist six when you add it all, an the last which is 2581=2 because the only number which have value its 8, and is it 2, if you dont get its for my explanation because my inglish its very bad, well i'd never been good in mathematics, but this took me five min to resolve it, I'm proud of that, jaja, at least I'm like six years old kid, jaja It's tricky because it's mislabelled.

It has nothing to do with "Math" in the sense we're thinking when we read "Math Problem". The solution isn't in mathematics, it's in the ability to recognize a pattern in images that, in this case, just happen to be numbers. That being said, it's fun and I plan on driving my friends batty with it. :) Abel I agree with those saying it is mathematics - it's not what you traditionally think of when you think "math", but just reading the comments, some people used a mapping, while others used topology.

The fact that it uses our symbols for numbers doesn't make it "not math", instead it seems like it was designed to throw people for a loop. Anonymous If pre-elementary schoolers can figure it out then there is no computations involved.

Therefore you should look at the representation of the number rather than what it represents. I am a Senior in college about to do my student teaching for teaching high school mathematics.

Lets just say it takes a crazy person like me to look at it non-mathematically. Misha, I think you have got the magic wand that has turned this exercise into an interesting discussion. Looking into the loops could only give us the "exact" solution, but looking beyond the discussion and interaction would inspire us to analyse problems in the world with different perspectives, realising that there are many problems that may have multiple solutions, though some may explain better than the others.

So, by just presenting with a problem - like this, you could inspire people to THINK. Would a Maths teacher be able to generate such an interest in discussion, by posting a school problem? This is wonderful. Congratulations. Anonymous For all those that said they were PhD students or masters students or whatever in mathematics, I claim the answer might as well be anything.

As you know questions such as "find the next number in the sequence" are always invalid, as a real sequence is any injection from N into R. That is a sequence is only truely determined if you provide an instructions on how to get to the next one. If this were an arbitary sequence the next number might as well be Pi, since we were not given any properties of the sequence this is a valid answer.

Anonymous Consider the following sequence... 2,4,6,... What's next? As was stated earlier it can be anything. Today I choose the next number to be 73. And here is a formula that makes it so... (-1/3)(x-2)(x-3)(x-4) + 2(x-1)(x-3)(x-4) + (-3)(x-1)(x-2)(x-4) + (73/6)(x-1)(x-2)(x-3) so.. when x=1 then f(1)=(-1/3)(1-2)(1-3)(1-4) + 0 + 0 +0 f(1)=2 and when x=2 f(2)=0 + 2(2-1)(2-3)(2-4) + 0 + 0 f(2)=4 and when x=3 f(3)=0 + 0 + (-3)(3-1)(3-2)(3-4) + 0 f(3)=6 and magically :) when x=4 f(4)=0 + 0 + 0 + (73/6)(4-1)(4-2)(4-3) f(4)=(73/6)(3)(2)(1) f(4)=(73/6)(6) f(4)=73 So we simply use the idea that any product with a zero in it will always be zero.

This has the nice effect of cancelling out our big ugly formula. A similar logic could be applied to the original sequence to make the answer anything we want it to be.

It is not a math problem. It is a viso-spatial problem that uses numbers. How the answer is obtained is to count the number of enclosed areas in each single digit. For example 8 has 2 enclosed regions. 9 and 0 both have only 1 along with 6 and 4. If you add up the number of enclosed region per four digit "number" you arrive at the answer. For example 6764=3 because: 6 has one region 7 has zero regions 6 has one region 4 has one region summed together there are 3 enclosed regions. Not mathematics other than adding.

For reference I am a junior in EE and have taking maths up to and including Calculus 3 Anonymous I'm coming to the party rather late but wanted to express what a great exercise this was. It certainly had a few of us at work performing mental gymnastics. It is surprising how hard it can be for some of us to deliberately think or perceive differently. Thanks for posting it. I have to admit, I was a little taken aback by the posturing of some commenters and imagine it relates to either bruised egos, blind arrogance or lack of awareness of how people other than themselves think.

I think I understand those in the latter camp in that a right brain thinker has great difficulty in perceiving the way a left brain thinker would, and vice versa. Similarly, I might be able to wrap my mind around the mentality of someone whose arrogance prevents them from imagining approaches other than theirs might be worthy of consideration.

But I have great difficulty in understanding the apparent anger some posters have expressed, as if they had somehow been victimized by either the puzzle, the solution or the comments.

Great discussion on whether this qualifies as math also. It's amazing how some reject it outright, which might explain their difficulty solving the problem, since they have difficulty in adapting their thinking. Typical authoritarian follower responses from those who like their world in a tight package, always constant and reliable, never to be interpreted as anything but their literal definition of it.

Poor souls, really, as they will never discover anything new, just new ways of applying the status quo.


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