Results 1 to 6 of 6
Like Tree2Thanks
  • 2 Post By masters

Thread: Two homework problems

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    38
    Awards
    1

    Question Two homework problems

    I am having problems with these ones:

    The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.

    Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Henderson's Avatar
    Joined
    Dec 2007
    Posts
    127
    Thanks
    2
    I am having problems with these ones:

    The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.
    If it is divisible by 36, it is also divisible by 4, meaning the last two digits, 5B, must be divisible by 4 (B is 2 or 6).

    Also if it is divisible by 36, it is also divisible by 9, so A+5+5+B is divisible by 9, so A+B must be either 8 or 17.

    Since B is at most 6, it is impossible for A+B to equal 17, so the sum must be 8.

    Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?
    \frac{\frac{1}{2}+\frac{1}{3}+x}{3} = 1

    \frac{1}{2}+\frac{1}{3}+x = 3

    x = 2\frac{1}{6}

    Thank you!
    You're welcome, though there is a button for that type of thing (hint, hint...)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,550
    Thanks
    15
    Awards
    1
    Quote Originally Posted by fecoupefe View Post
    I am having problems with these ones:

    The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.

    Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?

    Thank you!
    The number A55B must be even and the sum of its digits must be divisible by 3. Assign B=0,4,6,8 and assign A in such a manner that the sum of all 4 digits is divisible by 3 and you arrive at the number: 6552 or 2556.
    -------------------------------------------------------------------------

    \frac{\frac{1}{2}+\frac{1}{3}+x}{3}=1

    \frac{1}{2}+\frac{1}{3}+x=3

    x=3-\frac{1}{3}-\frac{1}{2}

    x=\frac{13}{6}

    Edit: You beat me Henderson, but that's ok. I'll give you your thanks!! Hint. Hint.
    Last edited by masters; Oct 25th 2008 at 09:25 AM.
    Thanks from Henderson and TheMathQueen
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    2

    Done!

    1) For the first question I cant think of a particular method apart from brut force. U know its divisible by 36, therefore it must be in the 36 times table. You know it is a number that contains four elements. So just list all the numbers that fit this category. i.e. 1008-9976.

    Ul find that the numbers that come of the form A55B, are 2556 & 6552. Therefore in both cases A + B = 8!

    2) There is a much more methodical approach to this problem. For the average to be equal to one, you need the sum of all 3 numbers to = 3. so far u have 1/2 and 1/3 as the numbers. you need to put them under a common denominator, so 1/2 = 3/6 and 1/3=2/6. So remember you need the total to be 3 = 18/6. So your remaining number is simply 18/6 - 5/6 = 13/6!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2016
    From
    US
    Posts
    2

    Re: Two homework problems

    I have a question, though. Why must the sum of its digits be divisible by 3?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    15,073
    Thanks
    3169

    Re: Two homework problems

    Quote Originally Posted by TheMathQueen View Post
    I have a question, though. Why must the sum of its digits be divisible by 3?
    actually, the sum of its digits must be divisible by 9 (which also makes it divisible by 3)

    Why Divisibility Rules Work
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help with these homework problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Aug 4th 2010, 12:10 PM
  2. Need 3 Homework problems
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Nov 8th 2009, 07:50 AM
  3. Homework Problems
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 11th 2009, 12:24 PM
  4. 2 Homework Problems
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Nov 24th 2008, 09:19 AM
  5. Homework HELP!! 3 problems!
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Apr 13th 2008, 07:11 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum