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A Number Made Up Of Two Digits

  1. Jul 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increadsed by 9. Find the number.

    2. Relevant equations

    I think this will definitely involve simultanous, quadratic or both of the equations.

    3. The attempt at a solution

    Let the two digits number be xy.
    The sum of the digits is 11
    -----> therefore x+y = 11
    If the digits are interchanged, the original number is increased by 9
    -----> forming equation from this last part is where am stuck.
    From my own point of view, "if the digits are interchanged", means, yx instead of xy as in the first equation.

    I also understand the original number which is increased by 9 as the sum x+y = 11. If 11 is increased by 9 when the two digit are interchanged, then the second equation should look like this
    yx = 11+9
    then the two set of equation from the first and second sentence should look like this:
    x+y = 11
    yx = 20
    but this set of equations does not in any way help me, rather is making the problem more complex because the equation are not factorizable, each time I subtitute one variable into another such that quadratic equation is formed. I have used quadratic formular and completing the square method; I keep getting different values for x and y which is not true if you subtitute the values into the equations. Please I need help, maybe you can start by explaining what the second statement " if the digits are interchanged, the original number is increased by 9", mean and how to form equation with it. Thank you.
     
    Last edited: Jul 8, 2012
  2. jcsd
  3. Jul 8, 2012 #2

    phinds

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    Re: A Number Made Of Two Digits

    I think you are REALLY going about this the hard way. How many possible 2-digit numbers are there where the two digits add up to 11 ?

    It took me 6 seconds to find the answer
     
  4. Jul 8, 2012 #3
    Ok, this is how it goes:

    let x be tens digit and y be ones.
    The original number is 10x+y
    The reversed is 10y+x

    Sum is 11

    So

    x+y=11
    10y+x=10x+y+9

    then just go ahead and solve!
     
  5. Jul 8, 2012 #4
    That's where you went wrong.
     
  6. Jul 8, 2012 #5
    As I told I cannot form an equation from the second sentence because I do not understand it. I only understand the first part of the sentence.
    As for your question, the digits are, 47, 56, 92, 83, . They are four possible two digit numbers that add up to 11.
     
  7. Jul 8, 2012 #6
    Thanks for your concern, I think I have really over worked myself. Let me go to bed now, I will work over it later in the morning when I wake up. Thanks!
     
  8. Jul 8, 2012 #7
    Imagine it like this. In essence, when reversed, the number is bigger than the original.

    Let R be reversed number. In order for the reversed to be bigger than the original, you need to add something to the original. In this case, you add 9. So,

    R=O+9

    Now, it just so happens that we are not dealing with the numbers as a whole, but as individual digits. To form a number from digits x and y, we multiply each by its place (you may have learned this as expanded notation). So to form a number with digits x and y, the expression is 10x+y. Reversed, it is 10y+x. Then just substitute those into the above equation.
     
  9. Jul 8, 2012 #8

    phinds

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    Ah, well that's where you went wrong. There are several more. For example 74 and 65
     
  10. Jul 9, 2012 #9

    phinds

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    Mod note: I edited out a complete answer that was quoted here.

    Zondrian, I see you are pretty new to the forum. Perhaps you have not yet gotten the idea but it is frowned on here to just spoon-feed an answer. The point is to help people learn how to GET the answer, not give it to them. Took me a while to get that myself, so no foul, but please try it that way in the future.
     
    Last edited by a moderator: Jul 11, 2012
  11. Jul 9, 2012 #10
    This is the working:
    the sum of the number which equals 11
    -----> x+y = 11
    the original number is 10x + y
    if it is interchanged, it becomes
    10y + x
    the two set of the equations now becomes
    x + y = 11 ------> (1)
    10y + x = 10x + y + 9
    10y-y + x - 10x = 9
    9y - 9x = 9
    y - x = 1 ------> (2)
    from equation (2)
    y = 1 + x
    putting y in equation (1)
    we have x + 1 +x = 11
    2x + 1 = 11
    2x = 11 - 1
    2x = 10
    x = 10/2
    x = 5
    x in equation (1)
    5 + y = 11
    y = 11-5
    y = 6
    see that 5 + 6 = 11
    and and that 65 = 56 + 9
    Thanks for your tips. But I must ask, apart from this method, is there no other method to use?
     
  12. Jul 9, 2012 #11

    HallsofIvy

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    You were given two methods in the responses here! One method was to use the two equations as given in your last post. The other was to recognize that the only pairs of one digit numbers that add to 11 are 2+ 9, 3+ 8, 4+ 7, and 5+ 6. Reverse those and see which satisfy the second condition.
     
  13. Jul 9, 2012 #12

    phinds

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    First of all, +1 on what HallsofIvy said.

    If you are inclinded to pure math, the equation method is good although to my mind excessively tedious for such a trivial problem. If you are, like me, only interested in SOLVING the problem, you should pay more attention to what I did:

    possibilities
    0 --- doesn't work
    1 --- doesn't work
    29
    38
    47
    56
    65
    74
    83
    92

    And the answer just jumps right out at you as 56/65 and as I said, it took 6 seconds. My POINT in this is that if you care about ANSWERS you should not get hung up on methodology.
     
  14. Jul 9, 2012 #13
    Thanks, I apreciate every bit of your effort?
     
  15. Jul 9, 2012 #14
    But if I must ask, what made you think that 10x + y is the original number?
    Can't we use say, 8x + 3y, 9x + 2y, 5x + 6y or any other number such that sum of it digit is 11. What made you think that 10x + y is the most suitable? I realy want to know why.
     
  16. Jul 9, 2012 #15
    As many of you that proposed 10x + y as the original number, what made you think that 10x + y is the original number?
    Can't we use say, 8x + 3y, 9x + 2y, 5x + 6y or any other number such that sum of it digit is 11. What made you think that 10x + y is the most suitable? I realy want to know why.
     
  17. Jul 9, 2012 #16

    HallsofIvy

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    Are you serious? You understand that our number system is "base 10" don't you? The number "19" has digits 1 and 9 and 19= 10+ 9. The number "35" means 3*10+ 9 and has digits 3 and 5. The number "ab", where a and b are digits, means a*10+ b.
     
  18. Jul 9, 2012 #17
    I took note of what you are saying, but I must inform you that am preparing for a future mathmatics exam. In the exam, we are required to form equation from word problems before solving so as to earm more marks. Ok!
     
  19. Jul 9, 2012 #18
    Thanks for that lecture, I never knew that this kind of rules apply when solving word problems such as this.
     
  20. Jul 9, 2012 #19

    phinds

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    Uh ... because we do our math in base 10? Do you really not understand positional notation?
     
  21. Jul 10, 2012 #20

    HallsofIvy

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    Basic definitions always apply!
     
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