Impossible Math Puzzle

[NunoMouraz]

1+1=3 IF you dont use a condom

[sydneyxfaircloth]

try all numbers evenly divisble by 7 . but not evenly divisble by 2 , 3 , or 4 .

49/2=24 r 1 49/3=16 r 1 49/4=12 r 1 49/7=7 r 0

its quite simple guys (:

there may be more solutions; hint , no even number will work .

[Bahzad23]

why are you guys thinking this so hard, just go to walmart and check how many cards are in a pack and thats your answer....duh

[JsKingBoo]

Here's a paradoxal math problem:

Does Infinity = Infinity - 1 or Infinity > Infinity - 1?

At first it may seem obvious, but think about it very carefully.

[TheYellowDK]

49, 133, 217, 301, 385, 469, 553, 637, 721, 805, 889, 973, 1057, 1141, 1225, 1309, 1393, 1477, 1561, 1645, 1729, 1813, 1897, 1981, 2056, 2149...

you just start with 49 and add 84 each time

[SolarFlare42]

The number of cards can be: 49, 133, 217, 301, ...

One way of expressing the formula is:

N = (7 + 12m)*7

where N is the number of cards, and m can be zero or any positive integer [0, 1, 2, 3, ...] There are an infinite number of answers.

[whoneedspapercutz]

It's easy to find the first answer, which is 49. It's finding the actual formula for this problem that gives me a hard time.

So far I got: 49 and 133.

But those numbers won't be good enough tell I find the formula code. Thanks for math problem, I havn't had to actually "think" since I got this new wannabe "algebra" crap for school.

[whoneedspapercutz]

It's easy to find the first answer, which is 49. It's finding the actual formula for this problem that gives me a hard time.

So far I got: 49 ,133 and 85.

But those numbers won't be good enough tell I find the formula code. Thanks for math problem, I havn't had to actually "think" since I got this new wannabe "algebra" crap for school.

[manudude02]

@whoneedspapercutz 85 isn't divisible by 7

[whoneedspapercutz]

@MrCerialKiller fail

[nuovaguardiapchdinam]

49...because 49/2=24(cards in each stack => 1 remaining) 49/3=16(cards in each stack=> 1 remaining) 49/4=12(cards ...=>1 remaining)...but 49/7=7 cards in each stack, none remaining...this is the smallest solution...there are bigger ones...but i've already find an answer, so no nead for another one:D

[gsaharan1990]

even exponential powers of 7.....

Eg.- 7^2=49, 7^4=2401...... and so on

that was so freaking easy

[KrisBlueNZ]

BTW you don't need the "stacks of 2, one left over" part.

If n-1 is an integer multiple of 4, it must also be an integer multiple of 2.

[TheAnonymousParadox]

Solutions are of the form 84k+49 for any k where k is an integer. Of course one can not have a negative number of baseball cards, therefore in this particular scenario

there are an infinite number of solutions for k=0,1,2...

When k=0, there are 84 x 0 + 49 = 49 baseball cards.

[noex100]

133! The answer is 133! The other answer is 49!

[yancee24LOL]

very easy. it's 49. the last sentence says it all.

[testalwaysdotcom]

watch?v=ZlFwLBAnpj0

[yjdte]

49. That was kind of easy.

[ChoujiAkimichi]

49, 133, 217, 301, 385, etc etc.

how did i find this? by cheating

made a program in c++ that looped with an incrementing variable checking that i % 2 == 1, % 3 == 1, % 4 == 1, and % 7 == 0.

:P

[brandonmcmahan]

I figured 49, just cause seven divides into it evenly, and all the others can divide into it with one left over.

[vivekpros]

49 is the answer and probz no other solutions.

[kakamadrid92]

@vivekpros what about 2401?

[deminisher98]

It is 49 cards because

49/2= 24R1

49/3= 16R1

49/4= 12R1

49/7= 7

[deminisher98]

@deminisher98 btw, this was done before I read any comments

[Agent2471]

49 cards

[MegaAsians]

7,49,77,91 etc.

i just thought of the multiplication table got all the odd numbers(clearly, if its an even number, theres no extra card if you divide the pile in two)

and if i get a number thats not divisible by 2,3 or 4 , thats my answer.

i'm too lazy to do the solutions or what u call them, my teacher is 'surprised' by my way of figuring things out

[sdrawkcabgnipytmi]

Your way is wrong.

49 is correct, your other solutions aren't.

For example 7/4 is1 remainder 3.

All solutions have to have a remainder of 1 when divided by 2, 3 and 4.

[Fredy216B]

it can't be seven because if he made a stack of four he'd have three left

[kingcow111]

119

[nhojmabon]

I don't understand mods, so here is my alternative solution: 2a=3b=4c=x-1 where x = number of cards Let a = 6n, b = 4n, c = 3n Find the integer n: Using equation (12n+1)/7 = Integer, all integer values of n can be tested to see whether it produces an integer or not. Values of n and the corresponding number of cards are: n=4, x = 49 n=11, x = 133 n=18, x = 217 n=25, x = 301 n=32, x = 385 n=39, x = 469 n=46, x = 553 .... And so on

[joepdevallis]

My math teacher made me do that one a week ago!!

[xAciidHackx]

??????

[kamakis22]

easy...

[TyYann]

let x the number of cards. x=n1×2+1 x=n2×3+1 x=n3×4+1 x=n4×7 hence x-1 is a multiple of 2, 3, 4 and x is a multiple of 7. So x-1 is a multiple of 12. The smallest x satisfying this is 49. And the other ones are 49 modulo 3*4*7=84. x=49 mod 84 Simple modular arithmetics. I like it.

[DeeDoubleYou81]

Nice work! Great explanation too.

[Thehamsterofmuppets]

I think its 49.

[DeeDoubleYou81]

You thought right! Can you find any other answers?

[J1man81]

it depends on how many stacks u have

[DeeDoubleYou81]

so are you saying there is more than one answer? if you are, you are right, but do you know the answers?

[acdc10133]

oh i came up with 721 but i did it quite lazily, i guess there's a bunch more. i just did 2x3x4x5 and added one=120 so you'd use 121 and add 120 until it worked :/

[DeeDoubleYou81]

hmmmm... that is one way to do it!

[SuperTooonlink]

The lowest correct answer is 49, then u can add any amount of 84 and it still is correct

133, 217, 301, etc...

[DeeDoubleYou81]

You got it! Nice work. :)

[shawnalori44]

that was so easy it made me want to cry x)

its 49,derderder xD

[DeeDoubleYou81]

Let's not forget the second half of the problem "is there more than one solution?"

I'd wait before you starting crying ;)

[arthritismutilans]

I fear that looking at all of these equations, nobody will visit this page again

[DeeDoubleYou81]

No way! This is too awesome! Thank you for sending all this info to me as a message, that way I can hang on to it!

And thanks for all your time in figuring it out!

[arthritismutilans]

OK. let me do it via here

e1= s ( n1n2n3/ n1)

r(3) + (4 x7)s= 1 gives r=-9, s= 1 and e1=4x7x1=28 the original remainder of this is 1, ae= 28

r(4) + (3x7)s = 1 gives r=-5, s= 1 and e2= 3x7x1=21 the original remainder of this is 1, ae=21

r(7) + (3x4)s= 1 gives r= 7, s=-4 and e3= -48 the original remainder of this is 0, ae= -48 x0= 0

[arthritismutilans]

hence

ae1 + ae2 + ae3 = 28+21+0=49

The solution will occur at a modular of 3x4x7=84, hence 49+n(84)

can you see what it is doing?

let's say for equation 1:- original form y= 1 + n(3)

[arthritismutilans]

e= 4x7s guarantees that it can be divided by 4 and 7 with no remainder, but since

e1 = 1- r(3), the remainder is 1 when e1 is divided by 3,

to get a remainder of 1 in the original equation of y=C1 +n(3), you multiply e with C1( which is 1), hence C1e1= s(4x7)x1 when s =1, c1e1=28

[arthritismutilans]

equation 2:- original form y= 1+ n(4)

e= s(3 x7) guarantees it can be divided by 3 and 7, with no remainder, but since

e2 = 1- r(4), the remainder is 1 when divided by 4,

to get a remainder of 1 in the original equatiohn of y= C2+n(4), you multiply e with C2 ( which is 1), hence C2e2 = 1 (3x7)x1=21

Do the same for equation 3

[arthritismutilans]

now canyou see what happens when you add all c1e1+ c2e2+c3e3=y

consider the situation when y is diided with 3, c1e1 will ensure it gives a remainder of 1, c2e2 and c3e3 is completely divided by 4 and 7 with no remainder

if y divided by 4, c2e2 ensures a remainder of 1, but c1e1 and c3e3 is completely divided by 4 with no remainder

similar thing will happen for division with 7, c3e3=0 will ensure a remainder of 0 and the other two are divisible by 7

[arthritismutilans]

you will arrive at one of infinite solutions, in our case it just happened to be the smallest whole number, but we already know the answer will recur at 3x4x7=84, hence y= 49+n(84)

if you arrive at a y which is bigger than 49, you just keep subtracting 84 until you arrive at the smallest positive whole number i.e. 49

[DeeDoubleYou81]

This sounds absolutely right and perfect! Thanks again!!

[arthritismutilans]

Can I give you a little puzzle in exchange then,

What shape must a rectangular box be to give maximum volume over surface area ratio?  An instinctive answer is the cube , but ..... prove it!!

[DeeDoubleYou81]

I have milled through this problem a few different ways and after some derivatives and manipulation.... my short answer is this...

You are asking to maximize the ratio of Volume to Surface Area, which I show as V/SA..

Let's first look at only the units..

cubed/squared which reduces to a linear function...

We both know a linear function has no maximum or minimum..

Thus, this particular problem has no defined solution...

[DeeDoubleYou81]

I think you might have meant to ask what will the dimensions of a box be in order to maximize volume and MINIMIZE surface area, which can be solved. ;)

[arthritismutilans]

that is what maximizing volume over surface area ratio means, V/SA is maximum, as you increase the volume there wil come a point where the surface area will also increase, thereby the V/SA is no longer optimal, if you reduce SA, so will the V reduce so that V/SA is no longer optimal. The quest is to prove that the most efficient shape is the cube. VA/SA will only reduce to a linear equation if all sides are equal, but now you have 3 unknown variables, so it does not reduce to a linear equation

[arthritismutilans]

Volume of cube / SA= 1/6a

let the three dimensions not be equal and a is the smallest of them all

a, ab, and ac denotes the three dimension

V/SA = 1/6 a ( bc/ b+d+bc)

since b and d>1 ( as a is the smallest side)

therefore bc/ b+c+bc cannot be more than 1, and the entire expression

1/6 a (bc/b+c+bc) cannot be more than 1/6a, i.e. the V/Sa is only maximal if the three sides are equal

Thus the solution, bye

[mikksters]

gosh math gives head aches =_='

[DeeDoubleYou81]

Damn.... I have worked this and along the way came up with 1/6a, but didn't think it was right. I have done and re-done this problem using derivatives and partial derivatives to find min and max. Here you simply do it. Geeze, sometimes I over think the most simple things!!

Thank you for posting the solution! Great problem!!

I remember doing a similar one in a calc class. It said "find the height of a cylindrical container so that the volume is maximized and the surface area is minimized."

[arthritismutilans]

hi tried replying to your cylinder problem, using the same approach, but it won't post

[arthritismutilans]

so much for my long equations essentially, there is no solution for h given r, but the maximum Va/Sf approaches 1/2

[arthritismutilans]

Unless, your question is on just minimizing surface area ( which does not make sense, let r or h =0, answer solved

or maximum volume ... let r or h is infinite, but Va/Sf in the cylinder is a bit strange

[DeeDoubleYou81]

Hey, sorry for the delayed response, I'm right in the middle of my move to Japan. I should have internet up at my new apartment soon and will see if I can find the solution in my old course work. I remember it being a shape similar to a tomato soup can so maybe h=4r.

[arthritismutilans]

now remainder of

a. divide by 3= remainder 1, hence ex1= 28

b. divide by 3= rem 1, hence ex1= 21

c. divide 7= rem 0, hence ex0=0

e1+e2+e3= 49

it will recur at a modulus of N=3x4x7=84

hence our solution y= 49+n(84)

[arthritismutilans]

I have just replied you ,but part of the solutio nis lost on youtube, I typed twice, has it gone into your accoutn

[DeeDoubleYou81]

Hmmm.... I only have what is above...

"hence our solution y= 49+n(84)"

which sounds absolutely right! you are awesome!!!

[arthritismutilans]

Then if you do the same for y2 and y3, and add them all up, you will find one of infinite solutions which satisfy all 3 equations. You then subtract n1n2n3 from that number until you reach the smallest whole number.

[arthritismutilans]

\the clever bit of the remainder theorem is that it finds an equation such that

y1 gives a remainder of 1 when divided by n1, but a remainder of 0 when divided by n2 and n3.

You then multiply y1 with C1 so that the equation gives a remainder of 1 x C1.

[DeeDoubleYou81]

I really think you are on to something here! I really wish you had a camera so I could check out your notes.

Do you think this method will give the same result as I found in my solution?

[arthritismutilans]

took me a while to understand this without modular arithmetic background but the solution is quite dissapointing, it does not avoid guessing answers, it only makes the quessing easier, but I never suspected we could avoid guessing either

[DeeDoubleYou81]

Hi there,

I took a look at the Chinese Remainder Theorem and followed it as far as I could (quite complex without modular arithmetic knowledge as you mentioned) and I came up with nothing. I agree there should be a mathematical method to solve this problem without guessing, but I'm not sure there is. Maybe my knowledge of abstract math is not adequate for this type of reasoning.

I really enjoy reading your solutions though! Keep 'em coming!!

[arthritismutilans]

hiya,

bascially you have to find the form

r1n1 + s (N/n1)= 1

and N= n1n2n3etc

let e= s(N/n1)

and one solution to the multiple equations is

y= sum of c1e1 + c2e2 +c3e3

the rest of the solutions will be modular of n1n2n3= N

[arthritismutilans]

Hi,

I have since come across something known as the Chinese Remainder theorem, solution to multple congruence, which hopefully will lay this problem to rest once and for all.

[arthritismutilans]

Wish some number theory experts could lend us a hand, I am sure this is a rather elementary problem to them, singingbanana on the web might be able to clarify. But till then, I have got a boring essay on How I spend my holidays to submit to my school teacher, bye

[DeeDoubleYou81]

I appreciate your explanation! You have really got the wheels in my brain a spinning, this is great! I feel for ya on that essay, no fun at all! When I was in school, all I cared about was math, I hated English. Later on, I learned how to enjoy writing and saw its importance. Thanks again for the insight!

GOOD LUCK!!

[arthritismutilans]

hahaha,

You probably did not go through it, but thanks for the sentiment. Are you a school teacher yourself, wearing a tie and in front of a white board. Or did you finally start to love writing when you started writing reports ( which are awfully dry), proposals or research thesis stuff etc. That's what my bro does. It's awful and only Maths is the true language, the universal language, the uncolored language.

[DeeDoubleYou81]

I went through the first bit that you did and WILL for sure go through the rest when I have more time (right now I'm preparing to move to Japan) and I really appreciate you taking time to share all this!!! I taught high school math this last spring semester and will be teaching English in Japan starting in the fall.

I started to love writing when I could finally write about subjects that interested me. I even like poetry, imagine that! :)

But you are very right, math is the ONLY true language!

[arthritismutilans]

They must satisfy the equation

7a-12b =1

where a and b are whole numbers

if you convert this into the form

y = mx +c

a= 12/7b + 1

draw the graph

where a and b are whole numbers, they will give you the solution,

The solution is 7a

[arthritismutilans]

The second limiting factor is from

12b = 7a -1

therefore

12b = ( sqrt 7a + 1) ( sqrt 7a -1)

since 12b is whole , sqrt 7a is also whole , i.e. 7a is a complete square, therefore, the solution can only exist as 49 ( 7x7), 98 ( 7 x2 x7 x2), hence a is restricted such that a= 7d squared where d is also whole, put that into equation 1

7d^2 = 12/7b +1

[DeeDoubleYou81]

Very interesting. You first equation is simply a line whereas this equation is quadratic. Now, both of these provide continuous solutions vs discreet solutions. I think the graph of the solution would be some sort of step function. I'd love to see a graph of your work. If you can, make a video response so we can see you work.

[arthritismutilans]

Hullo,

Me no camera.

If you draw the line, you will find that the first solution is 49 ( a =7),

Then next would occur at fix interval of 12a =84

In fact the linear equation proves that there can be no intervening whole number solution between 49 ( a=7) and 49 + 84. ( a= 7+12) and 49 + n (84)

[arthritismutilans]

Imagine if 49 is the first solution on the straight line and the next solution is 49 + 84, then all values of a and b in between will not be whole. Since the line is linear, this pattern will repeat itself periodically.

[arthritismutilans]

This actually raises an important generalization,

for a set of equations which demand whole number solutions where a, b, c are prime ( divide the cards by 2, 3, 5, etc) and C1, C2, C3( leftover of 1, 2, 3 etc)

y= a (P1) + C1

y= b ( P2) + C2

y= c (P3) + C3

the solution

is y= M + n( abc)

where M is the first whole number that satisfy all three equations and n is whole.

[arthritismutilans]

The great difficulty is of course finding the first number M that satisfy the triplet of equation.

in your case it is easy because C1 and C2 is the same, hence we can collapse both equations

y= 3 (P1) +1

y= 4 (P2) + 1

y = 3 (4) ( P) +1= 12P +1

The challenge is when C1, C2 and C3 are different, and you have more than one equation to satisfy

[arthritismutilans]

I can think of a cumbersome way of doing it, pair the equations up and solve them pair by pair,

hence

for

y= a(P1) +1

y= b(P2) +2 find the form y = M+ n1 ( ab)

y= c(P3) + 3

y= d ( P4) + 4 find the form y = m+ n2(cd)

Then, find the form

M-m = n1(ab) -n2(cd) gives you another linear equation to solve for n1 and n2

[CumicAnimations]

Dudes im just saying, i realised i diddnt actually rate it so now i put it to 1star. xD

[DeeDoubleYou81]

Wow, thanks man. You are amazing!!

[MattAtaka]

49 cards, cba with another answer lol

[DeeDoubleYou81]

Good so far!

[Shortmandesigner]

I got 49 and 91 as 2 possible answers

[Shortmandesigner]

whoops, nvm, not 91, missed the part about being in stacks of 4

[CumicAnimations]

Definatley not impossible but cool. 4 star

[freddygotfingered23]

yes the one who posted this and all of us knew that its not impossible look at the description lol

[DeeDoubleYou81]

Thank you! I'm glad you said it!

[Vicimentis]

Yes the first answer is 49. but there are infinite answers. its either the GCM or the LCM but its 84,which is the fir4st common multiple besides one, if you add 84 to 49, you will always get another solution. ie 49-133-217-301-385-469 etc.

[DeeDoubleYou81]

Absolutely, thanks for pointing that out!

[VitalSigns1]

Pretty easy if you ask me. 49

[DeeDoubleYou81]

Yes, that's right. But... Is there more than one answer? Can you figure out what they are?

[markyw19]

You have 49 cards...I kno bcuz in stacks of 2 with one left over it must be an odd number, it works for 3 , and 4, and its the second one i tried lol, as for the other possibilities i am not going to try more but i would assume there would be more, not sure tho