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Math Help - Scroboscopi

  1. #1
    Junior Member
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    Scroboscopi

    On each day of its life a scroboscopus square its number of legs. For example, if a scroboscopus has 2 legs on it first day of life, it would have 4 legs on its second day, and 16 legs on its third day. Tom bought some newborn scroboscopi. Some had 2 legs, some had 3 legs, and some had 5 legs. The total leg count was 58 legs. The next day the leg count was 164. The following day, the leg count was 1976. How many scroboscopi did Tom buy?
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  2. #2
    Senior Member DivideBy0's Avatar
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    Well, you could set up a group of simultaneous equations to be solved using either elimination or Cramer's Rule:

    Let x be the number of 2 legged scroboscopi
    and y be the number of 3 legged scroboscopi
    and z be the number of 5 legged scroboscopi

    2x + 3y + 5z = 58 (1)
    4x + 9y + 25z = 164 (2)
    16x + 81y + 625z = 1976 (3)

    Well, since I'm bored and I don't have/know how to use Latex I'll just go through elimination with you.

    Let's firstly eliminate the x, since it's the simplest to get rid of. Futhermore, we must use a combination of equations.

    Using the first 2 equations:
    2x + 3y + 5z = 58 (1)
    4x + 9y + 25z = 164 (2)
    (1) * 2
    4x + 6y + 10z = 116 (1')
    (2) - (1')
    3y + 15z = 48 (4)

    Then using the first and third:
    2x + 3y + 5z = 58 (1)
    16x + 81y + 625z = 1976 (3)
    (1) * 8
    16x + 24y + 40z = 464 (1'')
    (3) - (1'')
    57y + 585z = 1512 (5)

    Simultaneously solving (4) and (5) we get:
    3y + 15z = 48 (4)
    57y + 585z = 1512 (5)
    (4) * 19
    57y + 285z = 912
    (5) - (4')
    300z = 600
    *z = 2*

    Going back to (1) and (2):
    2x + 3y + 5z = 58 (1)
    4x + 9y + 25z = 164 (2)
    Plugging in z and simplifying we get:
    2x + 3y = 48 (1''')
    4x + 9y = 114 (2')
    (1)*2
    4x + 6y = 96 (1'''')
    (2) - (1'''')
    3y = 18
    *y = 6*

    Plugging y into (1''') we get:
    2x + 18 = 48
    2x = 30
    *x = 15*

    Yes I know my equation labels are messed up, anyway:

    ~-*There are 15 scroboscopi with 2 legs, 6 scroboscopi with 3 legs, and 2 scroboscopi with 5 legs*-~

    or something like that...i think
    Last edited by DivideBy0; March 31st 2007 at 07:13 AM.
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  3. #3
    lissa91
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    scroboscopus

    Is there any other way to solve the problem?
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  4. #4
    MHF Contributor
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    There are lots of ways to solve simultaneous equations. There are all equally tedious unless you get someone else to do all the arithmetic.

    For example, I typed into MathCad the 12 numbers from DivideBy0's equations, hit the magic inverse button, and out popped 15, 6, and 2. It took all of two minutes, including the time to load the porgram.

    Your calculator may have similar capabilities.
    Last edited by TKHunny; August 14th 2007 at 03:58 AM.
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  5. #5
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    Hello, aznmartinjai!

    On each day of its life, a scroboscopus squares its number of legs.
    For example, if a scroboscopus has 2 legs on it first day of life,
    it would have 4 legs on its second day, and 16 legs on its third day.

    Tom bought some newborn scroboscopi.
    Some had 2 legs, some had 3 legs, and some had 5 legs.
    The total leg count was 58 legs.
    The next day the leg count was 164.
    The following day, the leg count was 1976.
    How many scroboscopi did Tom buy?
    I solved it with an augmented matrix . . .


    Let B = number of bipeds.
    Let T = number of tripods.
    Let P = number of pentapods.

    We have the equations: . \begin{array}{ccc}2B + 3T + 5P & = & 58 \\ 4B + 9T + 25P & = & 164 \\ 16B + 81T + 625P & = & 1976\end{array}

    So we have: . \begin{vmatrix} \;2 & 3 & 5 & | & 58\; \\  \;4 & 9 & 25 & | & 164\;  \\ \;16 & 81 & 626 & | & 1976\; \end{vmatrix}

    \begin{array}{c} \\ R_2-2\!\cdot\!R_2 \\ R_3-8\!\cdot\!R_1\end{array}\;<br />
\begin{vmatrix}\;2 & 3 & 5 & | & 58\; \\ \;0 & 3 & 15 & | & 48\; \\ \;0 & 57 & 585 & | & 1512\;\end{vmatrix}

    \begin{array}{c}R_1-R_2 \\ \\ R_3-19\!\cdot\!R_2 \end{array}\;<br />
\begin{vmatrix}\;2 & 0 & \text{-}10 & | & 10\; \\ \;0 & 3 & 15 & | & 48\; \\ \;0 & 0 & 300 & | & 600\;\end{vmatrix}

    \begin{array}{c}R_1 \div 2 \\ R_2 \div 3 \\ R_3 \div 300\end{array}\;<br />
\begin{vmatrix}\;1 & 0 & \text{-}5 & | & 5\; \\ \;0 & 1 & 5 & | & 16\; \\ \;0 & 0 & 1 & | & 2\;\end{vmatrix}

    \begin{array}{c}R_1+5\!\cdot\!R_3 \\ R_2 - 5\!\cdot\!R_3 \\ \end{array}\;<br />
\begin{vmatrix}\;1 & 0 & 0 & | & 15\; \\ \;0 & 1 & 0 & | & 6\; \\ \;0 & 0 & 1 & | & 2\;\end{vmatrix}


    Therefore: . \boxed{\begin{array}{ccc}B & = & 15 \\ T & = & 6 \\ P & = & 2\end{array}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Hey, Divideby0 . . .

    Nice job!

    Great layout and good use of color . . . impressive!

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  6. #6
    Senior Member DivideBy0's Avatar
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    Thanks Soroban, my inspiration came from your posts of course
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