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Math Help - Sum of the factors

  1. #1
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    Sum of the factors

    Hi,

    I need some help in finding the sum of the factors of 18000 which are multiple of 8 but are not multiple of 25

    i proceeded as below.

    18000 = 2^4 * 3^2 * 5^3

    Now
    factors that are multiple of 8 = 2* 3^2 * 5^3
    factors that are multiple of 8and 25 = 2 * 3^2 * 5

    sum = sum[factors that are multiple of 8] - [factors that are multiple of 8and 25]

    where m i going wrong and how to proceed?

    Thanks.
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  2. #2
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    Hello, skyskiers!

    If they had asked for "the number of factors which are multiples of 8
    but not a multiple of 25", there are some theorems we could use.

    . . . There are 12.

    But they asked for the sum of these factors.
    I see no choice but to find all these factors and add them.

    It is not a very intellectual problem, is it?


    Find the sum of the factors of 18,000 which are multiple of 8 but are not multiple of 25.
    There are 60 factors.
    I found them by simple division.

    . . \begin{array}{cc}<br />
1 & 18000 \\ 2 & 9000 \\ 3 & 6000 \\ 4 & 4500 \\ 5 & 3600 \\ 6 & 3000 \\ {\color{red}8} & 2250 \\ 9 & 2000 \\ 10 & 1800 \\ 12 & 1500 \end{array} . . . \begin{array}{cc}15 & 1200 \\ {\color{red}16} & 1125  \\ 18 & 1000 \\ 20 & 900 \\ {\color{red}24} & 750 \\ 25 & {\color{red}720} \\ 30 & 600 \\ 36 & 500 \\ {\color{red}40} & 450 \\ 45 & 400 \end{array} . . . \begin{array}{cc} {\color{red}48} & 375 \\ 50 & {\color{red}360} \\ 60 & 300 \\ {\color{red}72} & 250 \\ 75 & {\color{red}240} \\ {\color{red}80} & 225 \\ 90 & 200 \\ 100 & 180 \\ {\color{red}120} & 150 \\ 125 & {\color{red}144} \end{array}


    As promised, there are 12 such factors.

    Their sum is: . 8+16+24+40+48+72+80+120+144+240+360+720 \:=\:\boxed{1872}

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  3. #3
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    Hello Soroban,

    Right , doesnt seem intellectual...probably that was the only reason i posted it here ..thought if there existed any better way...

    Thanks once again..
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  4. #4
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    Quote Originally Posted by skyskiers View Post
    Hello Soroban,

    Right , doesnt seem intellectual...probably that was the only reason i posted it here ..thought if there existed any better way...

    Thanks once again..
    There indeed is
    Any factor, with required criteria is of the form

    2^x.3^y.5^z

    where x \in \{3,4\}, y \in \{0,1,2\} and z \in \{0,1\}

    What do can you say about
    (2^3+2^4)(3^0+3^1+3^2)(5^0+5^1)

    Do you see the argument?
    Thanks
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    There indeed is
    Any factor, with required criteria is of the form

    2^x.3^y.5^z

    where x \in \{3,4\}, y \in \{0,1,2\} and z \in \{0,1\}

    What do can you say about
    (2^3+2^4)(3^0+3^1+3^2)(5^0+5^1)

    Do you see the argument?
    Thanks

    Yes Aman, got the argument thick and fine.... great logic... got me out of the trouble ....

    Thanks..
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