• Sharebar

So it seems that there are some math people around. To follow up on http://www.wallstreetoasis.com/forums/harder-brainteaser-than-the-coin-flip here is another one:

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

Show that:
a+b=c+d

Requires middle school math, but sure can ruin an interview.
As before, SBs for 1st correct answer.

Edit - all numbers are strictly positive. Apologies for confusion, but the problem remains open.

Edit - To provide an update, this has, thus far, been proven in 1 confirmed way by nonTargetChimp 9 (see his text file proof), and also in 1 more way that appears correct by unForseen (or rather by a friend of unForseen that works in HR). Another proof using trigonometry has been proposed and has not been fully evaluated but could be correct, and blastoise tried to prove this using fancy linear algebra but so far that proof does not appear to be correct, since he was able to prove it ignoring the positive constraint, meaning he proved something that is false. The most elegant solution to date, in my opinion, would have to be the HR rep's as presented by unForseen, which is found on page 8.

Background on the problem - came from an 8th grade city level math competition from the late 1960s at a mathematical school in the USSR.

Comments (399)

  • In reply to sinthushan
    Flake's picture

    sinthushan wrote:
    Just divide the 2nd equation by the first it's really simple.

    (a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) a+b=c+d

    Fail.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • TonyPerkis's picture

    Take the square root of each variable?

    I eat success for breakfast...with skim milk

  • Flake's picture

    My guess (disclaimer: I am not mathematically inclined):

    Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

    We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • In reply to Flake
    Dr Joe's picture

    Flake wrote:
    My guess (disclaimer: I am not mathematically inclined):

    Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

    We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.


    Your intuition may be correct but it is certainly not a proof.

  • In reply to Dr Joe
    Flake's picture

    Dr Joe wrote:
    Flake wrote:
    My guess (disclaimer: I am not mathematically inclined):

    Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

    We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.


    Your intuition may be correct but it is certainly not a proof.

    I fail as well.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • In reply to Flake
    Dr Joe's picture

    Flake wrote:
    Dr Joe wrote:
    Flake wrote:
    My guess (disclaimer: I am not mathematically inclined):

    Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

    We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.


    Your intuition may be correct but it is certainly not a proof.

    I fail as well.


    SB for the chuckle

  • UFOinsider's picture

    WHOOPS, edited...

    a^2 + b^2 = c^2 + d^2

    SO

    a^2 + b^2 = a^2 + b^2

    AND

    a^3 + b^3 = a^3 + b^3

    THEREFORE

    a^x b^x = a^x b^x

    SO

    a^1 b^1 = a^1 + b^1

    AND

    a + b = a + b = c + d

    a + b = c + d

    Or something like that. Apparently, a 5th grader is smarter than me.

    Get busy living

  • lookatmycock's picture

    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3

    just guessing:

    FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
    SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

    substitute right side of first equation into left side of 2nd equation:
    (a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

    divide both sides by (c^2 + d^2),

    and end up with (a+b) = (c+d)

  • In reply to lookatmycock
    Flake's picture

    lookatmycock wrote:
    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3

    Is my proof correct?

    FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
    SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

    substitute right side of first equation into left side of 2nd equation:
    (a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

    divide both sides by (c^2 + d^2),

    and end up with (a+b) = (c+d)

    My GMAT level math tells me that's...kind of not right.

    (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2) does not equal a^3 + b^3 = c^3 + d^3.

    I also don't get UFO's thing and how that proves a + b = c + d.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • In reply to lookatmycock
    FutureBanker09's picture

    lookatmycock wrote:
    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3

    just guessing:

    FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
    SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

    substitute right side of first equation into left side of 2nd equation:
    (a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

    divide both sides by (c^2 + d^2),

    and end up with (a+b) = (c+d)

    a^3 + b^3 does not equal (a+b)(a^2 + b^2)
    (a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3

  • In reply to Flake
    lookatmycock's picture

    Flake wrote:
    lookatmycock wrote:
    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3

    Is my proof correct?

    FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
    SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

    substitute right side of first equation into left side of 2nd equation:
    (a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

    divide both sides by (c^2 + d^2),

    and end up with (a+b) = (c+d)

    My GMAT level math tells me that's...kind of not right.

    (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2) does not equal a^3 + b^3 = c^3 + d^3.

    I also don't get UFO's thing and how that proves a + b = c + d.

    function yoFuckThisShit(www, forum, topic, x)
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    do while x > 0
    {
    yoFuckThisShit('WSO', 'monkeyingaround', 'Do You Know Algebra?', x)
    msgBox.Display ('This shows college and grueling engineering psets didn't prep me for shit', alert);
    x++
    }

    }

    init_app()
    {
    this.yoFuckThisShit ('WSO', 'monkeyingaround', 'Do You Know Algebra?', 1);

    }

  • Dr Joe's picture

    The above two corrections are correct (and necessary). Also, all material here is within the scope of the GMAT, so if you prefer, think of it as a data sufficiency question:

    Is a+b=c+d?
    (1) a^2+b^2=c^2+d^2
    (2) a^3+b^3=c^3+d^3

  • Will Hunting's picture

    Should I really embarrass everyone by doing this right now?

    I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls

    "Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

  • In reply to FutureBanker09
    Flake's picture

    FutureBanker09 wrote:
    C,

    let a = 1 b = -1 c = 1 d = 1
    a^2+b^2=c^2+d^2 holds true
    a+b=c+d is not equal

    let a = 1 b = 1 c = 1 d = 1

    a^2+b^2=c^2+d^2 holds true
    a+b=c+d is equal

    1 is insufficient

    same logic applies to 2

    Only when combining can you lock down the sign so C

    That's what I was trying to say in my post with the whole plugging of the numbers thing and picking positive/negative signs.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • lookatmycock's picture

    somebody go do it the long brute way, if this isn't a data sufficiency question:

    first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
    then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

    and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

    second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

    plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

    of course there's a shorter way but im an idiot today :D

  • In reply to FutureBanker09
    trazer985's picture

    FutureBanker09 wrote:
    lookatmycock wrote:
    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3

    just guessing:

    FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
    SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

    substitute right side of first equation into left side of 2nd equation:
    (a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

    divide both sides by (c^2 + d^2),

    and end up with (a+b) = (c+d)

    a^3 + b^3 does not equal (a+b)(a^2 + b^2)
    (a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3

    no it doesnt.

    a+b(a^2 + b^2) = a^3 + ba^2 + ab^2 + b^ 3

  • In reply to lookatmycock
    Dr Joe's picture

    lookatmycock wrote:
    somebody go do it the long brute way, if this isn't a data sufficiency question:

    first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
    then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

    and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

    second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

    plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

    of course there's a shorter way but im an idiot today :D


    plug all that crap in, solve, then plug in some more and see what you get

  • In reply to Dr Joe
    lookatmycock's picture

    Dr Joe wrote:
    lookatmycock wrote:
    somebody go do it the long brute way, if this isn't a data sufficiency question:

    first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
    then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

    and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

    second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

    plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

    of course there's a shorter way but im an idiot today :D


    plug all that crap in, solve, then plug in some more and see what you get

    shhhhh. im trying to make someone go through all that.

  • maximumlikelihood's picture

    Let me give it a shot.

    First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].

    Now, we know that a^3 + b^3 = c^3 + d^3, and so

    (a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)

    For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).

    If a+b = c+d, then

    (a+b)^2 = a^2 + 2ab + b^2
    (c+d)^2 = c^2 + 2cd + d^2

    But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.

    I don't know about you guys, but I have a life + work to do so if its wrong, so be it.

    Just Do It

  • In reply to maximumlikelihood
    Dr Joe's picture

    maximumlikelihood wrote:
    Let me give it a shot.

    First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].

    Now, we know that a^3 + b^3 = c^3 + d^3, and so

    (a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)

    For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).

    If a+b = c+d, then

    (a+b)^2 = a^2 + 2ab + b^2
    (c+d)^2 = c^2 + 2cd + d^2

    But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.

    I don't know about you guys, but I have a life + work to do so if its wrong, so be it.


    Well you are the first one on the right track. You have shown that proving a+b=c+d is equivalent to proving ab=cd, which is correct. However, in proving 2ab=2cd, you implicitly assumed a+b=c+d, hence you proved that if a+b=c+d, then a+b=c+d, which is redundant. However, thanks for bringing a hint of actual math to this thread.

  • maximumlikelihood's picture

    I'm pretty sure for the next part I have to cross multiply the two equations and show that ab = cd, but man what the hell I don't have time for this shit

    Sorry about the redundant proof, I felt bad typing it in

    Just Do It

  • maximumlikelihood's picture

    Also, I just showed that if a+b = c+d, I will get this result because

    (a+b)^2 = (c+d)^2, and a^2 + b^2 = c^2 + d^2.

    Just Do It

  • In reply to maximumlikelihood
    Dr Joe's picture

    maximumlikelihood wrote:
    I'm pretty sure for the next part I have to cross multiply the two equations and show that ab = cd, but man what the hell I don't have time for this shit

    Sorry about the redundant proof, I felt bad typing it in


    The reason I posted this is that the actual proof is quite elegant and non-trivial, and cross-multiplying doesn't come close to cutting it. It's a good problem to solve in boring classes.

  • Will Hunting's picture

    I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

    "Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

  • In reply to Dr Joe
    Will Hunting's picture

    Dr Joe wrote:
    Will Hunting wrote:
    I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

    Factoring will not be sufficient here. You keep saying you can do it - we are all waiting with SBs in hand.

    I think he meant factorization. Anyway, I will solve it once I get out of class and have time to type.

    "Look, you're my best friend, so don't take this the wrong way. In twenty years, if you're still livin' here, comin' over to my house to watch the Patriots games, still workin' construction, I'll fuckin' kill you. That's not a threat, that's a fact.

  • Flake's picture

    Dr. Joe you're such a tease.

    Under my tutelage, you will grow from boys to men. From men into gladiators. And from gladiators into SWANSONS.

  • pszonkadonk's picture

    This is probably stupid, but I go to a nontarget so give me abreak.

    take the natural log of everything thus:

    2lna + 2lnb = 2lnc + 2lnd

    divde by 2

    lna + lnb = lnc +lnd

    raise e^ln. thus

    a + b = c + d

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