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Math Problem

  1. Jul 30, 2004 #1
    i recently sat a national mathematics competition and came across this question, which i thought was pretty good. im posting it here to see how long it takes everybody to figure out. (the question has been wordedly differently)

    The sum of three children's ages is 23. The product of their ages is 113 more than the product of their ages this time last year. What is the sum of the squares of the children's ages? (easy after you figure out their ages)

    enjoy...
     
  2. jcsd
  3. Jul 30, 2004 #2

    Galileo

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    I got 259.
     
  4. Jul 30, 2004 #3
    Yeah it is 259. It took me a minute or two less then ten.
     
  5. Jul 30, 2004 #4
    yep, good work. most of the other people who sat the test moved past it because it would take too long and didn't go back to it. it did take me a little while, probably around 5 minutes, but it was an important question.
     
  6. Jul 30, 2004 #5
    This actually took me only 5 seconds because I have heard it from my classmates many times, I am bored with it ...:tongue2:
     
  7. Jul 30, 2004 #6
    lol... would have helped if i had known it going into the test.
     
  8. Jul 30, 2004 #7

    Gokul43201

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    Much easier if you don't try to figure out their ages.

    It took me a couple minutes to find the sum of squares, but I still don't know any of their ages.
     
  9. Jul 30, 2004 #8
    wow....i did it a weird way then.
     
  10. Jul 30, 2004 #9
    Yes, I read it.
    I am sorry I didn't mean it was 100% the same as those I have heard, but the fact is that in such exams, people like to give their students a long long long question to make them unable to concentrate on the main thing which in this problem it "sum of the squares of the children's ages", and other things will be irrelevant
    You already have x1+x2+x3=23 right ?
    But what if you were asked to find out each person's age ? How many minutes will it take you before you jump up and down in happiness "Binga Binga Binga I got it" ? -lol

    By the way, you should remember to take heart medicines before stepping into the room. I read your thread about your sm@rtie-ness and I really admire you as always, true.
    If you had another heart attack, people should worry about you much more...and that would also bring you bad evaluation then...-lol :tongue2:
     
  11. Jul 30, 2004 #10

    Gokul43201

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    [tex] x+y+z=23[/tex]

    [tex] xyz = 113 + (x-1)(y-1)(z-1) = 113+ xyz - xy+yz+zx + x+y+z - 1[/tex]

    [tex]So, ~~(xy+yz+zx) = 113 +23 -1 = 135[/tex]

    [tex]But,~~(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)[/tex]

    [tex]Hence,~~x^2 + y^2 + z^2 = 23^2 -270 = 259 [/tex]
     
    Last edited: Jul 30, 2004
  12. Jul 30, 2004 #11
    lol. i don't have a clue how i figured it out then, but i definitely remember finding the ages before squaring them to find what the question asked for. that heart thing truly stuffed up my calc exam...just really bad timing. btw i dont generally shout 'binga' :P im more the 'eureka' type :D. just kidding. thanks for the math Gokul. its strange to have someone admire me....that i don't know. thank you though. farewell.
     
  13. Jul 30, 2004 #12
    If i didn't do it that way, how could I say I could solve that problem in five seconds ?
    It is impossible to deal with this problem in such a short time.
    Don't ever say farawell please, whenever you and I come here, we are still able to meet again, that Forune makes us meet...-lol
    My bino should have some problems but anyway, you sound fun, like Jimmy-p
     
  14. Jul 31, 2004 #13
    The sum of three children's ages is x. The product of their ages is y more than the product of their ages this time last year. What is the sum of the squares of the children's ages?

    x*x + 2(1-x-y)
     
  15. Jul 31, 2004 #14
    LOL, i did this test as well
    i i dun it wrong though
    i took a guess,
    b.185....

    Heres another one from the test.
    i kinda got confused but friends after told me the answer.
    (note: this wasnt hard, just the way it was written, it really confused me, also its not word for word either)

    John tells truth Monday, Tuesday, Wednesday,and Thursday. and lies every other day. Jack tells truth Friday, Saturday, Sunday and Monday. one day both said "YesterdayI Lied". the day they both lied was....
     
    Last edited: Jul 31, 2004
  16. Jul 31, 2004 #15

    Gokul43201

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    I hope one of the choices was "none of the above" ...

    Hey, davilla, nice to see you back ! Wot's bin up witcha ?
     
  17. Aug 2, 2004 #16

    NateTG

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    It's pretty clear that they don't ever lie on the same day. However, it's also clear that today is Friday, so I suppose that the answer that they're looking for is Thursday.
     
  18. Aug 4, 2004 #17
    nope, the answer is Friday
     
  19. Aug 4, 2004 #18

    NateTG

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    Clearly Jack tells the truth on Friday, therefore Friday cannot be a day that they both lied. As you have written it, the problem asks for the day that both lied ,which does not exist since, on each day of the week at least one of the boys tells the truth. Since the problem refers to both saying, "Yesterday, I lied" it's possible to conclude that the statement is made on Friday, but that isn't what it's asking for. It is, however possible to infer that "the day that they lied" refers to the day that they claim that they lied on which would be Thursday.
     
  20. Aug 4, 2004 #19

    Gokul43201

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    I think this is just a question of interpretation or rather, correct misinterpretation.

    If the question is "on which day did they both say "yesterday I lied" ?" then the answer is friday.

    If it's "Which day did they both claim to have lied on ?" the answer is thursday.

    Finally, if the question is really "On which day did they both lie?" then the answer is 'on no day' !
     
  21. Aug 4, 2004 #20
    In the first problem with the ages, I found the ages are 3, 5, and 15. (I used a program to figure this out.)

    How would I go about figuring this out with the information given?
     
  22. Aug 6, 2004 #21

    Gokul43201

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    From post #10 use the following :
    #1. sum = 23
    #2. sum of squares = 259
    #3. sum of pairwise products = 135, an odd number.

    First sum = odd means you must have (a)1 odd + 2 evens OR (b)3 odds. Now, case(a) can be eliminated by using fact #3, since in that case, all the pairwise products (oe, oe, ee) will be even.

    So, we are looking for 3 odd numbers.

    Next use the fact that the for a given sum (or perimeter of cuboid) the sum of squares (or length of body diagonal) is minimized when the numbers are equal (when the object is a cube).

    So, in this case given that sum =23, if the numbers were roughly equal, the sum of squares would be less than 200 (64+64+49 = 177), but this is actually much larger (259). So we must consider extremely unequal numbers. So one of the following cases must be true : (i) 1 number much less than 8 and 2 numbers greater than 8, OR (ii) 1 number much greater than 8 and 2 numbers less than 8. Now we can eliminate most of case (i) through the following reasoning :

    sum of squares of the 2 larger numbers < 259...so the only allowed pairs are (9,9), (9,11), (9,13), (9,15), (11,11). All other pairs are eliminated. It's simple to go through these 5 cases and see that none of them work.

    Now you are left with case (ii). Clearly, the maximum value of the largest number is 15 since17^2 = 289 > 259. And since we want an extreme case, we start with (15,y,z) followed by (13,y,z) and (11,y,z) - this last case being pretty unlikely. For the first case, y+z = 23-15 = 8. So possibilities are (1,7) and (3,5). Clearly, the first pair fails but the second works.

    So, were are through...the solution is (15,5,3).

    In your code, you would have x = 1..21, y = 1..22-x , z = 23 - x - y - a total of 210 computations. Notice that we had to make only 6 computations (of wrong sets) before finding the correct set.
     
    Last edited: Aug 6, 2004
  23. Aug 7, 2004 #22
    Thanks Gokul43201, I was struggling with that for a bit.
     
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