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Help on a few puzzle problems

  1. Jun 4, 2004 #1
    These we're the problems that i've tried to do. Solutions with steps would greatly be appreciated.

    1. Suppose that a, b, c, are distinct non-zero digits.
    (A) Find a formula, depending on a,b, c, for the sum of all six digit integers whose only digits are a,b,c.
    (B) What is the sum of all six digit integers having no non-zero digits?

    2. Verify that 9^2+2^2=7^2+6^2 by cutting a 9x9 square into four pieces that together with a 2x2 square can be rearranged to give a 7x7 square and a 6x6 square.

    3. A point within an equilateral triangle is at distance 5, 7, 8 from each of the vertices. What is the length of a side of the triangle?

    4. A photographer wished to arrange twelve people of different heights in two rows of six each. Each person in the first row must be shorter than the person directly behind and, going from left to right, the people must get taller.
    How many such arrangements are there?

    5. Let S be a set of positive integers, each of which is less than 20.
    (A) What is the maximum size m of S if S has the property that no sum distinct elements of S is equal to 20.
    (B) Of all sets S of positive integers less than 20 having maximal size m, and having the property that no sum of distinct elements of S is equal to 20, which are the sets having the samllest possible sum?

    Thanks
     
  2. jcsd
  3. Jun 4, 2004 #2

    AKG

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    1. Suppose that a, b, c, are distinct non-zero digits.
    (A) Find a formula, depending on a,b, c, for the sum of all six digit integers whose only digits are a,b,c.


    Well, there are [itex]3^6[/itex] different possible 6-digit numbers. If that six-digit number is expressed DEFGHI, then 1/3 of those 6-digit numbers will have a in the D spot, 1/3 will have b in the D spot, and 1/3 will have c. The same is true for all other spots. So, the answer is:

    [tex]3^5(111,111)(a + b + c)[/tex]

    (B) What is the sum of all six digit integers having no non-zero digits?

    What? Are you sure you asked this question right? If it has "no non-zero digits" then it has all zero digits. The sum would clearly be zero.

    2. Verify that 9^2+2^2=7^2+6^2 by cutting a 9x9 square into four pieces that together with a 2x2 square can be rearranged to give a 7x7 square and a 6x6 square.

    Just try a few things yourself. I doubt this requires any special education (the first one didn't really either), just try it out till you get it. If anything, that's all I would do anyways, so you may as well do it yourself.

    4. A photographer wished to arrange twelve people of different heights in two rows of six each. Each person in the first row must be shorter than the person directly behind and, going from left to right, the people must get taller.
    How many such arrangements are there?


    The person in the front left must be shortest, and the person in the back right must be tallest. So, there is only one way to fill those spots. Now, you have 10 spots to fill, and 10 people. Now, the second tallest must be in either the back left or the front, second from the left, so there's 2 ways to place him/her. I'm sure something similar can be said of the 11th tallest. Now, if you put the 2nd tallest person in the back left, then the third tallest must go in the front, second from left. If you put the 2nd tallest in the front, second from left, then you must put the third in either the back left or the front, third from left. Again, something similar can be said for the 10th. As you can see, if I continue, this will get too complicated, but you have a start, I suggest you try to finish the rest.

    Another possible approach: look for a pattern. If you have 2 people, there is 1 possible arrangement. With 4 people, there are 2. With 6 people there are 5. With 8 people, see if you can figure it out, then post what you get, i.e. finish this off:

    1, 2, 5, _

    If a pattern emerges, you might be able to use to hypothesize a general formula, then prove it inductively.

    5. Let S be a set of positive integers, each of which is less than 20.
    (A) What is the maximum size m of S if S has the property that no sum distinct elements of S is equal to 20.


    I'm assuming all elements in S must be distinct. I'm guessing the set S = {10,11,12,13,14,15,16,17,18,19} fits the bill, where m = |S| = 9. And we can prove that 10 or greater won't work, because if we take any number, X, then we must eliminate 20-X as a possible element of the set, otherwise X + 20-X = 20. Choosing more than 10 elements means, eliminating more than 10 other elements (more than 20 in total), but there are only 19 elements to choose from (positive integers less than 20 are 1 through 19), so its impossible.

    I might look at the other problems later.
     
  4. Jun 5, 2004 #3

    Gokul43201

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    Hmmm...I think you're making a mistake here.

    Also, I think every odd numbered (in height order) person has 2 possible places (exept the first guy), while every even person has only one place to take. So the answer would then be 2^5 or 32 ways.
     
  5. Jun 5, 2004 #4

    AKG

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    That's definitely not true:

    2--- 8--- 9--- 10-- 11-- 12
    1--- 3--- 4--- 5--- 6--- 7

    7--- 8--- 9--- 10-- 11-- 12
    1--- 2--- 3--- 4--- 5--- 6

    2, an even numbered guy, has 2 places to go. Based on this, there are 3 places 3 could go. If 2 is in the top situtaion, 3 must be where he is. If 2 is in the bottom situation, 3 must be either where he is now, or where "7" is. Anyways, your answer may be right but the approach is wrong, as far as I can see.
     
  6. Jun 5, 2004 #5

    Gokul43201

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    Clearly, it is. Perhaps I needed some sleep.
     
  7. Jun 7, 2004 #6

    uart

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    I've got a slightly "brute force"ish solution to this one using just the cosine rule. Denoting the side length as "x" and the interior angles from the said point to the vertices as "q1" and "q2" (the third angle of course being 2Pi-q1-q2), gives the following set of three equations in three unknowns.

    x^2 = 74 - 70 cos(q1)

    x^2 = 113 - 112 cos(q2)

    x^2 = 89 - 80 cos(q1 + q2)


    If you re-arrange the first two equations for cos(q1) and cos(q2) and then substitute these into the third equation while replacing the cos(q1+q2) with "cos*cos-sin*sin" and also replacing the sin terms with sqrt(1-cos^2) then you can get it to a 4th order polynomial in x^2. I didn't solve it because it looked tedious, but I'm sure it could be done that way.
     
    Last edited: Jun 7, 2004
  8. Jun 7, 2004 #7

    matt grime

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    there is an elegant result involving equilateral triangles that states the sum of distances from an interior point to the sides is constant, hence the side is 20 units.
     
  9. Jun 7, 2004 #8

    Gokul43201

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    That IS nice (assuming you mean "distance to vertices")...but would mean that the side is 10 units, not 20.

    2x = l1 + l2 + l3, if you move the interior point onto a vertex.
     
    Last edited: Jun 7, 2004
  10. Jun 7, 2004 #9

    uart

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    Interesting result Matt, I haven't seen that one before. But the "distance to sides" referred to is surely the perpendicular distance to the side and not the distance to the vertices as in this problem. So I don't really see how it applies.

    The result x=20 (or x=20/sqrt(3) which is what I think you meant) is not consistent with the cosine rule equations above. I don't think either of those is the solution.
     
  11. Jun 7, 2004 #10

    Gokul43201

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    I think uart's right there. The distances to verticaes is not a constant as can be seen from the Toricelli construction of the Fermat point.

    Here's another "brute force-ish" approach. Draw the lines from the interior point to the vertices, as well as perpendiculars to each of the 3 sides. So you have 6 right angled triangles. There are now 7 unknowns : the perpendicular distances d1, d2, d3; the side x; and the distances from the feet of the perpendiculars to the vertices, say a, b, c. Fortunately, you also have 7 equations : 6 Pythgorean relations and matt's relation, stating d1+d2+d3=x*sqrt(3)/2.

    Have fun !
     
  12. Jun 7, 2004 #11

    uart

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    Well for what it's worth I just ran a very quick and rough numerical solution to my above cosine rule simultaneous equations and I'm pretty sure the approx solution is, x = 11.3578, correct to 4 decimal places.
     
  13. Jun 7, 2004 #12

    Gokul43201

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    Hmmm...11.3578 = sqrt(129), correct to 4 decimal places. So you had
    x^2 = 129, eh ? I wonder if there's some (symmetric) numerical route (of dimension x^2) from 8,7,5 to 129. sum of squares is 138...difference is 9...(not 7 or 14) ?
     
    Last edited: Jun 7, 2004
  14. Jun 7, 2004 #13

    uart

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    Ok Gokul, I just plugged sqrt(129) (to 12 decimal places) into the equations and it was correct (to within the precision used). So it looks like sqrt(129) is indeed the solution.
     
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