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Math Help - challenge questions

  1. #1
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    challenge questions

    1.
    A number has three digits and is equal to 12 times the sum of its digits what the number is?

    Solution:

    Assume three digit number = xyz, then

    100x + 10y + z = 12(x + y + Z)

    100x + 10y + z = 12x + 12y + 12Z

    Solve to get

    88x = 2y + 11z

    Take x=1, then y=0 and z= 8 in order to satisfy the equation.

    Three digit number is 108, hence 108 = 12(1+0+8).

    Is there any other way to solve this problem?

    2.

    The sum of first N integers is a three digit number with all of its digits equal, what is the value of N?

    1,2,3,4,5,6,7. Is an arithmetic series

    S= n/2[2a + (n-1) d] = n (n +1) / 2

    n (n +1) / 2 = 111x

    Since sum of first integers is a three digit number with all digits equal
    Hence it xxx = 100x +10x +x = 111x.

    By trial and taking x=6, and solving for n, n2 +n 1332 = 0, I will get n=36, is there any other way of solving;
    n (n +1) / 2 = mod 111, and getting value of n=36 from this equation?

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  2. #2
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    Quote Originally Posted by nazz View Post
    1.
    A number has three digits and is equal to 12 times the sum of its digits what the number is?
    Solution:
    Assume three digit number = xyz, then
    100x + 10y + z = 12(x + y + Z)
    100x + 10y + z = 12x + 12y + 12Z
    Solve to get
    88x = 2y + 11z
    Take x=1, then y=0 and z= 8 in order to satisfy the equation.
    Three digit number is 108, hence 108 = 12(1+0+8).
    Is there any other way to solve this problem?
    NO. 2 equations, 3 unknowns.

    But you can "manipulate" such that answer becomes apparent:
    11z = 88x - 2y
    z = 8x - (2/11)y
    Only y=0 is possible...get it?
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  3. #3
    MHF Contributor
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    Quote Originally Posted by nazz View Post
    The sum of first N integers is a three digit number with all of its digits equal, what is the value of N?
    1,2,3,4,5,6,7. Is an arithmetic series
    S= n/2[2a + (n-1) d] = n (n +1) / 2
    n (n +1) / 2 = 111x
    Since sum of first integers is a three digit number with all digits equal
    Hence it xxx = 100x +10x +x = 111x.
    By trial and taking x=6, and solving for n, n2 +n 1332 = 0, I will get n=36, is there any other way of solving;
    n (n +1) / 2 = mod 111, and getting value of n=36 from this equation?
    Use quadratic; discriminant will be sqrt(1 + 888x);
    only 6 works; don't know if that's any faster...
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