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Thread: Variable base problem

  1. #1 Variable base problem 
    Forum Junior epidecus's Avatar
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    Had these problems on a competitive math assessment. These are the ones that stumped me, so I'd like some help in how to work them. (note: the test is already done and we are allowed to take the tests home for personal use. I'm not sure whether or not my next mathematics meet will discuss its problems or not, so here I am). Responses are much appreciated.

    >>>>
    When the number is written in base , the value is 341. When the number is written in base , the value is 44. Compute .
    >>>>

    I'm familiar with the concept of base systems, but I was never introduced to working them in equations. So I'm not sure where to start here. What methods do I need to be aware of, and how would I go about evaluating and ?

    >>>>
    You cut a 3x3 square from the page of a calendar. All the squares contain a date and the sum of the 9 dates is a multiple of 13. What number is in the lower left corner of the 3x3 square?
    >>>>

    The "date" is the day's number in the month. Not sure what to do here...


    Last edited by epidecus; October 31st, 2012 at 11:08 PM. Reason: typo and add in
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  3. #2  
    Forum Senior TheObserver's Avatar
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    Well if x^2 is 341 in base y then we know that 1*1 + 4*y + 3*y^2 = x^2. Likewise if y^2 is 44 in base x then 4*1 + 4*x = y^2. This is essentially how you make equations out of different base systems, hope that starts you off on the first problem.


    Last edited by TheObserver; October 31st, 2012 at 10:19 PM. Reason: oops
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  4. #3  
    Forum Junior epidecus's Avatar
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    Quote Originally Posted by TheObserver View Post
    Well if x^2 is 341 in base y then we know that 1*1 + 4*y + 3*y^2 = x^2. Likewise if y^2 is 44 in base x then 4*1 + 4*x = y^2. This is essentially how you make equations out of different base systems, hope that starts you off on the first problem.
    Thanks; from there it's a straightforward sys. of equations.

    But what did you do exactly? Do I have to form a polynomial with the digits as coefficients and the base as the variable?
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  5. #4  
    Forum Senior TheObserver's Avatar
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    Its the same for numbers in base 10 if you think about it. The number 84938 can be written in the form 8*x^4 + 4*x^3 + 9*x^2 + 3*x + 8*1 where x in this case is 10, as we dealing with a number in base 10. Any number in any base can be written like that. If abcd is a number written in base x, then to turn it into base 10 write it as the polynomial a*x^3 + b*x^2 + c*x + d*1. Im not explaining too well, Im sure someone here could give you a formal treatment on base systems.
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  6. #5  
    Forum Junior epidecus's Avatar
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    Well, I think that answer alone is at least enough to help me solve the problem.

    So from the start. and .

    Therefore, . And .

    From this system, I've derived four pairs of solutions. I'm guessing non-naturals don't make sense as base values, so the only pair that makes sense is 8 and 15. Giving me a sum of 23. Is this right?
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  7. #6  
    Forum Junior epidecus's Avatar
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    Oops, base subscripts should've been on the other side of the equality.

    Anyway, anyone have a clue as to how to do the second problem?
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  8. #7  
    New Member TheWalrus's Avatar
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    Quote Originally Posted by epidecus View Post

    Anyway, anyone have a clue as to how to do the second problem?
    I think I have it... but if someone else could check it that would be great

    Anyways, I set up my problem like this:

    1) I drew a 3x3 square
    2) I named the first square "x". The next square is x+1, the next is x+2, and this completes the first row of the 3x3 square. Then for the middle left square, it is x+7 because we are on a calender, so the date below a box is seven days from the initial date (the date below "x" is 7+x). The next two are "x+8" and "x+9". This completes the middle row. For the bottom row I did the same thing, getting: "x+14", "x+15", "x+16".

    When you add all these x's and constants, you get: 9x+72=13y

    I didn't know how to advance from here (if that's even possible) but by guess and check I got x=5 (which makes y = 9). If x is 5, then the bottom left of the 3x3 square is "x+14" which equals: 19
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  9. #8  
    Forum Junior epidecus's Avatar
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    Quote Originally Posted by TheWalrus View Post
    1) I drew a 3x3 square
    2) I named the first square "x". The next square is x+1, the next is x+2, and this completes the first row of the 3x3 square. Then for the middle left square, it is x+7 because we are on a calender, so the date below a box is seven days from the initial date (the date below "x" is 7+x). The next two are "x+8" and "x+9". This completes the middle row. For the bottom row I did the same thing, getting: "x+14", "x+15", "x+16".

    When you add all these x's and constants, you get: 9x+72=13y

    I didn't know how to advance from here (if that's even possible) but by guess and check I got x=5 (which makes y = 9). If x is 5, then the bottom left of the 3x3 square is "x+14" which equals: 19
    Sounds like the right method to me, and it checks out. Clever!

    I don't really see a shortcut way to a solution though. The only practical numbers for and would be natural numbers less than or equal to 31. But I'm not sure how we could incorporate that into the equation so we could directly derive a solution. So I guess guess-and-check is the way to go, unless someone else knows a more efficient approach.

    Oh, and welcome to the forum Walrus
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  10. #9  
    Forum Junior epidecus's Avatar
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    Okay I got it. --> -->

    Since the value of isn't really what we're looking for, and it's just some integer, we simply say .

    Since I'm multiplying by 9 and then dividing by 13, and the result is a multiple of 13, I can disregard the factor of 9. So I just have to find the number that satisfies , which evaluated for is of course 5. (this approach is a lot simpler when you're thinking the problem through in your head)
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