Results 1 to 8 of 8

Math Help - Problem

  1. #1
    Junior Member
    Joined
    Dec 2005
    Posts
    58

    Problem

    I need help to solve this problem:

    Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.


    I have represented that two-digit number as
    10x+y=3(x+y)

    Multiplying both sides with (x+y) we get that
    \frac{10x^2+11xy+y^2}{3}=(x+y)^2

    Dividing left side with 3 we will get that
    \frac{10x^2+11xy+y^2}{9}=(x+y)^2
    which is number that we want.

    So I think we must solve system of equation
    10x+y=3(x+y)
    \frac{10x^2+11xy+y^2}{9}=(x+y)^2

    I have tried to solve this equation by puting x=\frac{2y}{7} (from first equation) into second equation but the result is x=0 and y=0.

    Is my way of solution right?
    help to solve.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by DenMac21
    I need help to solve this problem:

    Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.


    I have represented that two-digit number as
    10x+y=3(x+y)

    Multiplying both sides with (x+y) we get that
    \frac{10x^2+11xy+y^2}{3}=(x+y)^2

    Dividing left side with 3 we will get that
    \frac{10x^2+11xy+y^2}{9}=(x+y)^2
    which is number that we want.

    So I think we must solve system of equation
    10x+y=3(x+y)
    \frac{10x^2+11xy+y^2}{9}=(x+y)^2

    I have tried to solve this equation by puting x=\frac{2y}{7} (from first equation) into second equation but the result is x=0 and y=0.

    Is my way of solution right?
    help to solve.
    You made a mistake, on step 3, your result should look like
    <br />
\frac{10x^2+11xy+y^2}{3}=(x+y)^2<br />
    You accidently did 2 steps in step 2. You multiplied by (x+y) and divided by 3, then you forgot that you divided by 3 and divided by 3 again.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by ThePerfectHacker
    You made a mistake, on step 3, your result should look like
    <br />
\frac{10x^2+11xy+y^2}{3}=(x+y)^2<br />
    You accidently did 2 steps in step 2. You multiplied by (x+y) and divided by 3, then you forgot that you divided by 3 and divided by 3 again.

    No, I divided by 3 in step 3 because the square of sum of digits is equal to triple number which is wanted. So I didn't have to divide it by 3 but later it should be divided.

    The problem is that I can't find values of x and y.

    I can't solve this system of equation:
    <br />
10x+y=3(x+y)<br />
    <br />
\frac{10x^2+11xy+y^2}{9}=(x+y)^2<br />
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by DenMac21
    No, I divided by 3 in step 3 because the square of sum of digits is equal to triple number which is wanted. So I didn't have to divide it by 3 but later it should be divided.

    The problem is that I can't find values of x and y.

    I can't solve this system of equation:
    <br />
10x+y=3(x+y)<br />
    <br />
\frac{10x^2+11xy+y^2}{9}=(x+y)^2<br />
    Since x \in \{1,2,3,4,5,6,7,8,9\} and y \in \{0,1,2,3,4,5,6,7,8,9\},
    trial and error shows that the first of your equations has solution:

    x=2,\ y=7,

    but this is not a solution of the second of your equations.

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by DenMac21
    I need help to solve this problem:

    Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.
    Let Z be the number wanted. The sum of the digits of the
    two-digit number is 9 as the digits are 2 and 7. So the last
    condition of the problem translates to:

    3Z=9^2=81,

    so:

    Z=27.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by CaptainBlack
    Let Z be the number wanted. The sum of the digits of the
    two-digit number is 9 as the digits are 2 and 7. So the last
    condition of the problem translates to:

    3Z=9^2=81,

    so:

    Z=27.

    RonL
    Thats ok, but we have get that x=2 and y=7 by guessing numbers.
    Is there algebraic way to get that numbers?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by DenMac21
    Thats ok, but we have get that x=2 and y=7 by guessing numbers.
    Is there algebraic way to get that numbers?
    Not by guessing, exhausive search over the ranges involved is not guessing.
    And if you don't like exaustive search you can try the following:

    10x+y=3(x+y) \Rightarrow 7x=2y,

    but 7 and 2 are prime and so y is a multiple of 7,
    and as 7 is the only single digit non-zero multiple of 7
    we have y=7 and the rest follows.

    ( y has to be non-zero, as if it were zero x would be zero,
    which is not permitted).

    RonL
    Last edited by CaptainBlack; January 31st 2006 at 03:21 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by CaptainBlack
    Not by guessing, exhausive over the ranges involved is not guessing.
    And if you don't like exaustive search you can try the following:

    10x+y=3(x+y) \Rightarrow 7x=2y,

    but 7 and 2 are prime and so y is a multiple of 7,
    and as 7 is the only single digit non-zero multiple of 7
    we have y=7 and the rest follows.

    ( y has to be non-zero, as if it were zero x would be zero,
    which is not permitted).

    RonL
    Thanks, nice solution.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum