Results 1 to 12 of 12

Math Help - Problem Solving

  1. #1
    Junior Member
    Joined
    Feb 2008
    From
    Victoria
    Posts
    39

    Problem Solving

    Please help. Not sure if this is a trick question. What is the sum of the digits of all 2-digit numbers between 10 and 99. Please explain
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by not happy jan View Post
    Please help. Not sure if this is a trick question. What is the sum of the digits of all 2-digit numbers between 10 and 99. Please explain
    Have you learned how to set up sums using this symbol? \Sigma?

    The way to do this would be to find the sum from 1 to 99, and then subtract the sum from 1 to 9.

    In sigma notation, this would be expressed as \sum_{k=1}^{99}k-\sum_{k=1}^{9}k

    Note that \sum_{k=1}^n k=\frac{n(n+1)}{2}

    Can you try to take it from here?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,970
    Thanks
    1011
    Quote Originally Posted by not happy jan View Post
    Please help. Not sure if this is a trick question. What is the sum of the digits of all 2-digit numbers between 10 and 99. Please explain
    This is an arithmetic sequence.

    The sum of terms in an arithmetic sequence is given by

    \frac{n}{2}(a + z) where a is the first number of the sequence, z is the last number of the sequence, and n is the number of terms.

    In this case, a = 10, z = 99, n = 90

    So this sum is...

    \frac{90}{2}(10 + 99)=45 \times 109 = 4905
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,970
    Thanks
    1011
    Quote Originally Posted by Prove It View Post
    This is an arithmetic sequence.

    The sum of terms in an arithmetic sequence is given by

    \frac{n}{2}(a + z) where a is the first number of the sequence, z is the last number of the sequence, and n is the number of terms.

    In this case, a = 10, z = 99, n = 90

    So this sum is...

    \frac{90}{2}(10 + 99)=45 \times 109 = 4905
    Oops, I think I read your question wrong. Are you finding the sum of the numbers, or the sum of all the digits of the numbers?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    From
    Victoria
    Posts
    39
    You lost me. Thats way beyond my ability
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,970
    Thanks
    1011
    Quote Originally Posted by not happy jan View Post
    You lost me. Thats way beyond my ability
    If you're finding the sum of the numbers, all you need is the first and last numbers and the number of numbers.

    There are 90 numbers in between 10 and 99, so n = 90.
    The first number is 10, so a = 10
    The last number is 99, so z = 99.

    Then just put your a, n and z values into the formula and simplify.

    But if you're finding the sum of all the digits, that's another story entirely...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2008
    From
    Victoria
    Posts
    39
    I need the sum of the digits
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by not happy jan View Post
    Please help. Not sure if this is a trick question. What is the sum of the digits of all 2-digit numbers between 10 and 99. Please explain
    Oops...misread the problem.

    Note that if we add all the digits together, we can generate a pattern...

    1+0+1+1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1+9

    =10+(0+1+2+3+4+5+6+7+8+9)={\color{blue}10}+{\color  {red}45}

    Now, if we keep going on, we see that

    2+0+2+1+2+2+2+3+2+4+2+5+2+6+2+7+2+8+2+9

    =20+(0+1+2+3+4+5+6+7+8+9)={\color{blue}20}+{\color  {red}45}

    We start see a pattern now.

    In the end, we will end up with {\color{blue}10+20+30+40+50+60+70+80+90}+9({\color  {red}45})=10(45)+9(45)=19(45)=\color{red}\boxed{85  5}

    I hope you can follow this!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,970
    Thanks
    1011
    Quote Originally Posted by not happy jan View Post
    I need the sum of the digits

    OK, well each group of 10 numbers will end in 0, 1, 2, 3, ..., 9.

    There's 9 groups of these, so there's 9 x 1, 9 x 2, 9 x 3, ..., 9 x 9.

    There's also 10 more 1's (because the first digit of the first 10 numbers is 1), 10 more 2's, 10 more 3's ... 10 more 9's.

    So that makes 19 x 1 + 19 x 2 + 19 x 3 + ... + 19 x 9, or if you like,

    19 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

    =19 \times \frac{9}{2}(1 + 9) = 19 \times 4.5 \times 10 = 855

    Hope that helped
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Prove It View Post
    OK, well each group of 10 numbers will end in 0, 1, 2, 3, ..., 9.

    There's 9 groups of these, so there's 9 x 1, 9 x 2, 9 x 3, ..., 9 x 9.

    There's also 10 more 1's (because the first digit of the first 10 numbers is 1), 10 more 2's, 10 more 3's ... 10 more 9's.

    So that makes 19 x 1 + 19 x 2 + 19 x 3 + ... + 19 x 9, or if you like,

    19 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

    =19 \times \frac{9}{2}(1 + 9) = 19 \times 4.5 \times 10 = 855

    Hope that helped
    I like your way.

    Did you use arithmetic sequences?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,551
    Thanks
    542
    Hello, not happy jan!

    Instead of asking "What's the formula?", we can use a primitive approach.


    What is the sum of the digits of all 2-digit numbers from 10 and 99?
    Imagine adding up the 90 two-digit numbers . . .


    \text{ten 1's} \begin{Bmatrix}1&0 \\ 1&1 \\ 1&2 \\ \vdots \\ 1&9\end{Bmatrix}\;\;\text{0 to 9 = 45}

    \text{ten 2's}\begin{Bmatrix}2&0 \\2&1\\2&2 \\ \vdots \\2&9\end{Bmatrix}\;\;\text{0 to 9 = 45}

    . . \vdots. . . . . \vdots . . . . . . \vdots

    \text{ten 9's}\begin{Bmatrix}9&0 \\ 9&1 \\ 9& 2 \\ \vdots \\ 9&9 \end{Bmatrix}\;\;\text{0 to 9 = 45}


    In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times.
    Sum of the units digits: . 9 \times 45 \:=\:405

    In the tens-column, we have: . \text{ten 1's + ten 2's + ten 3's + . . . + ten 9's}
    . . \;=\;10(1) + 10(2) + 10(3) + \cdots + 10(9) \;=\;10(1 + 2 + 3 +\cdots + 9) \;=\;10(45)
    Sum of the tens digits: . 450


    Therefore, the sum of the digits is: . 405 + 450 \;=\;{\color{blue}855}

    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,970
    Thanks
    1011
    Quote Originally Posted by Chris L T521 View Post
    I like your way.

    Did you use arithmetic sequences?

    --Chris
    I sure did, once I figure out there were 19 sums of the first 9 digits, I used the formula for arithmetic series

    S_n=\frac{n}{2}(a+z)

    Where n is the number of terms, a is the first term of the sequence, and z is the last, to work out the sum and then multiplied by 19.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help solving a problem.
    Posted in the Algebra Forum
    Replies: 0
    Last Post: August 13th 2011, 06:37 AM
  2. Further Problem Solving
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 18th 2010, 03:42 AM
  3. Problem Solving
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 19th 2008, 04:56 PM
  4. Need help in solving a problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 18th 2008, 04:02 AM
  5. Problem Solving
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 17th 2008, 08:25 AM

Search Tags


/mathhelpforum @mathhelpforum