Solution to Impossible Math Puzzle

[dorigzz]

2+2 brought me here

[anonimnystefy]

who said that this was impossible???whoever said this is impossible must be pretty dumb.

[PhilcoreTheLoud]

This actually is pretty cool. I`ve always been kind of on a warpath with math haha but I have to be good at it now for my studies and I find that puzzles like this help me a lot to raise my interest and hence my willingness to get involved with numbers more :D

[immortalhidan66]

22/7 and pie are different. On the iPod touch pie is 3.141592653589793(and more) and 22/7 is 3.142857142857143( and more)

[nerfmanz1]

smarty

[juanduism]

pretty kool indeed



[Twizzzle]

This isn't a proof. You're demonstrating properties of the concept but you haven't proven anything :P. Interesting though :D

[dlfite10b]

ok a simple question what is 22 / 7 (pie) and that should be the exact value

[Irig209]

I am going to show my kids this, thank you.

[Misguidance]

That was awesome!!

[Warpollo]

1110 hahah just read it vertical

[101animations]

i have proof that there isn't infinite numbers, thanks to a proffessor layton puzzle for this.

this equation is true with only one set of numbers

((X+1)=((Y+1)x1.5))=(X=(Yx2))

this only work where X=2 and Y=1

if number are infinite, then this should work with a fraction of infinite sets of numbers, which is of course, infinity, but this only works with the stated values

[kamehapa]

@101animations a subset of an infinite amount is not necessarily an infinite amount... That's like saying because X=Y=2 should work for a subset of numbers, therefore should have infinite solutions but only works for X=2 and Y=2. Sounds pretty silly when you say it like that though, right?

[vanstraten11]

So, does this solve the US debt problem????

[FlinckShinesOn]

.....Why did I enter the internet today?

Seriously ._.

[allusernamesaregone0]

There are more but.... yea, if you want to find the ones in the middle all you must do is multiply or divide

[allusernamesaregone0]

Then all you must do is go through the exponents of 7, or multiply it, to find other answers, which is simple. As he did. It's a pattern.

[allusernamesaregone0]

Out of letters last comment so... Then no 2 5 or 8 in the second 10s of 30s, and no 1s, 4,s 7,s or 0s of the last set. So the 4's are like 2s but you can count for the 4's. So what I did was find that 4 is what I can use. I counted up to see where it can have 1 leftover where 2, and 3 would follow. I got up to 48 at one point where everything followed. Then I checked with the 7 and voila. All this in my head. I don't understand how this is "impossible" when extraordinarily simple.

[allusernamesaregone0]

That was the easiest problem I've ever done, as in 7th grade advanced math. Simplify the problem. 2,3,4, and 7 is what you deal with. 2,3 and 4 have a multiple 1 behind, as 7 completes with none leftover. Obviously since 2 has 1 leftover it can't be even. So every even multiple of 7 can't be. 14, 28,42,56 and so on. Then 3's, simple enough, it can't be a multiple of 3. If you can easily count up to 30 with 3's. It should be very easy to figure out it can't be 3 6 or 9 in the first set of the 30.

[hannahbanana24510]

My teacher can do this one math trick where you take 2 diff numbers and both numbers are 3 digits and both numbers have 3 diff digits. Then you subtract the 2 numbers with biggest on top. Then, you tell her the number in either the ones or hundredths place and she can get it right every single time in a heartbeart and i wanna know how to do it.

[pokemonkiller20]

an easyer way to write it is 49 to the nth power.

[manmeester]

get a life

[pokemonkiller20]

@manmeester dud what is your problem. Some people like math. Do you know that people can get paid to do math problems. I bet you don't.

[manmeester]

@pokemonkiller20 go rip you of on pokemon maybe even they dont you want to do that

[pokemonkiller20]

@manmeester Your a dumb ass you know that. You just asume i like pokemon. Anyone with half a brain can tell that your an eight year old who thinks its cool to talk negitivly about everything. P.S. Start using correct grammer.

[robbiechinn]

This is how I did it

Say n is the number of cars

(n - 1) is divisible by 2, 3 and 4

so n = 12m + 1 where m is an integer

n = 12m + 1 = 13, 25, 37, 49, etc.

But it must be divisible by 7

So 12m + 1 = 49 + (7 x 12)p where p is an integer

(49 plus every 7th multiple of 12 will be divisible by 7)

So n = 49 + 84p

[junkyjuice21]

modular arithmetic and number theory. kinda cool

[chellylinnfarm]

is this a process of elimination and guess and check?

[guitardudeguy00]

You can solve this one in a top down process

[davidsonghello]

49~

[n2dbowser]

what grade math is the????????

[lmc6506]

nothing is impossible!

[wiijjimmy12]

you had 49 cards didnt yuo

[zamkam]

Your statement: "of course, numbers are infinite so there has to be an infinite number of answers" is nonsense. Some problems have no solution, some have a finite number of solutions, some have infinitely many, the one you chose happens to be of the latter kind, but in general it's very hard to find out which one is it. For all solutions to problems similar to the one you present here, look for the "Chinese Remainder Theorem" in any book on elementary number theory.

[devdevorah]

You could simply do 49+84a. Why is it "-n" instead of just "n"?

[yalg22]

nice vid . Im a math major and love these types of problems please post more.

[olio542]

i wish you were my math teacher...

my math teacher is a fat old lady that doesn't let you ask any questions, thanks to her im teetering between a "B" and "C"

[z0rn86]

why is this problem "impossible"?

it's quite simple..

[easy77146]

give me a hard q

[NosirrahKaraneeum]

So the answer is:

Any number equal to seven to the "X" power; where "X" divided by 2 has a remainder of zero.

Here's it written out:

N = quantity of the cards

X mod 2 = 0

(N = 7 ^ X)

Correct me if I'm wrong.

[DeeDoubleYou81]

You are correct! BUT, there are more answers. I am interested to see what you find for the general solution. :)

[crazyGodLikE]

35 = 11 howd u get 2 left over

[NosirrahKaraneeum]

35 divided by 3 is 11 with a remainder of 2.

[DeeDoubleYou81]

Yes, as NosirrahKaraneeum said, 35/3 is 11 r2. The problem states that you want only 1 left over, not 2. Keep trying :)

[Blaze468]

I love 'fun' math too. Especially ones that require you to use common sense/ thinking about it rather than just imputting numbers into a calculator. Kinda like the 7 girl w/ 7 big bags w/ 7 small bags w/ 7big cats w/ 7 small cats= how many legs

[DeeDoubleYou81]

Exactly! That "leg" problem is a good one!

[Kewickviper]

Very good explanation. I got the 49 in about 2 minutes just checking odd multiples of 7, but didn't persevere enough to get the GS. Kinda wished I hadn't looked at the answer straight away now since this is more fun than anything I do on my Maths degree.

[DeeDoubleYou81]

I'm glad you liked it! I love "fun" math, not so much the textbook math. When I taught it, I would always try to find interesting problems that the students could relate to, that way I kept their interest.

[enclave2k1]

I wonder if there's a more elegant solution to this and these types of problems. I was able to find the correct answer as you did, but I dont like the fact you have to enumerate x%7=0 initially, then finding the pattern manually.

I'd be interested in hearing your thoughts.

[DeeDoubleYou81]

There are other ways to solve this problem, in fact, a fellow youtuber named "arthritismutilans" laid it all out on the original problem video. He used the "Chinese Remainder Theorem" to solve it. Quite complex, but very impressive. Once you get an understanding of the theorem, it makes perfect since and seems much more efficient at finding the solution. You should check out his comments, they are from a while back.

[shikamarusgirl17]

LOL I hate school but I love logic problems and such so I guess Algebra isn't SO bad. XD I'm impressed at how you can have a formula for that sort of question's answer. It's amazing, and it never crossed my mind that there was any other way, besides listing multiples forever. Nice work!!

[DeeDoubleYou81]

Thank you! Yeah, you can pretty much use algebra to solve any problem. I'm glad you think algebra is okay. :D

Thanks for watching!

[Bella500000]

i'm in sixth grade

so i don't understand a word you said

BUT i think you got it because you sound smart and the math was great.

[DeeDoubleYou81]

Soon you will understand and maybe even better than me!

[nikonninja]

Getting a headache - think I will give this another look-see when I have a few shots lined up. :P Good job explaining btw.

[DeeDoubleYou81]

LOL, yeah it's a tricky one and shots always help in solving math problems ;)

Thanks for watching!!

[BookCollector100]

I've never been exceptionally gifted in mathematics.

I do enjoy looking at the processes of calculus mathematics in books, problems and methods, even though I cannot personally complete them.

[DeeDoubleYou81]

The more you look, the more you'll learn and one day, with enough effort, you will be able to solve them. :)

Math is a difficult subject for many, but I truly believe anyone can do ANYTHING if you put your mind to it!

[kb100]

Your solution set seems to be incomplete.

After some thinking i've come up with a general formula that solves your problem more completely.

S(n) = 7(12n + 7), where n is an integer 0 or greater, produces all solutions to the problem.

[DeeDoubleYou81]

Actually, if you look at my solution of

s = 7[7+(n-1)12]

and then simplify it as follows:

s = 49 + 84(n-1)

for n=1 to infinity, the first solution is 49.

Now yours is s = 7(12n+7) when simplified becomes:

s = 49 + 84n

for n=0 to infinity

By inspection, we have the same answer. The only difference is our initial condition, where mine is n=1 and yours is n=0. By convention, you start at n=1.

Thanks for working through it!

[singingbanana]

A good investigation. In general the problem can be solved using something called the Chinese Remainder Theorem (well, not exactly in this case because we had piles of 2 and then piles of 4 which messes up the theorem, but nothing we can't deal with).

[DeeDoubleYou81]

Very interesting, I never thought of this, I will have to look into it!

[arthritismutilans]

Good point actually, what happens when the modulus are not coprimal and they do not share the similar remainder. In this case it does, so all we do is to collapse the two equations into 1

y= 1mod(2) =1mod(4) but let

y1= 1 mod (2) and y2 = 3 mod (4) y3= 5mod (7)

lets see,

x= a1e1 + a2e2+ a3e3

a1e1 will be divisible by 4 and 7, but when divided by 2, won't yield 1, arrrrrgh, because if it is divisible by 4 then it should be divisible by 2

[arthritismutilans]

Can we group together all of the coprimal modulus, use the chinese remainder theorem on them, find a modular function solution and then solve it against the non-coprimal ones, i.e., group together 3, 4, and 7, solve for them and then only solve for the case of y= 1 mod 2

[arthritismutilans]

hence solving

r (4) + 7s =1 s=3, r=-5 e=21, ae=63

r(7) +4s =1 s=-5, r=3, e=-20, ae=-100

ae2 + ae3 =-37 mod (28) = 19 mod (28)

y2,3= 19+28n

y1 = 1+2n

but since 28 will definitely be divisible by the 1st modulus (2) because we used the 2nd modulus (4) in our theorem,

it falls on the constant 19 to fulfill our criteria, which in this case it does. Can I put forward a conjecture that as long as

[arthritismutilans]

we use the bigger modulus of the two non coprimal arrgh this doesn't work either it is just that this case is simple

[arthritismutilans]

19+ 28a= 1+2b

b-14a=9, the smallest whole number solution is actually a=0 and b=9, hence this solves our prob, y1= 1+ 2(9)=19, hurray!!!!

thanks for pointing me in the right direction, singingbanana

[singingbanana]

I think you have it.

[DeeDoubleYou81]

You guys rock!

[Vicimentis]

from 49, its always plus 84 ie 49-133-217-301-385-469 etc.

[DeeDoubleYou81]

I never thought of it like that, thanks for sharing!!

[markyw19]

wow very nice lol