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Math Help - Integers

  1. #1
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    Integers

    Dear all

    I have these questions i am having difficulties solving on my on own,your help is been sorted as to how to approach these questions and soultion.I pick these questions from schuam series.



    Thank you


    Question 1

    Of all integer pairs(x,y) that satisfy the equation 42x+55y=1,0nly one such pair has 100<x< 200.what is the value of x in this integer?

    Question 2
    Let X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 19 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)
    Last edited by barhin; May 19th 2011 at 12:33 PM.
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  2. #2
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    Quote Originally Posted by barhin View Post
    Question 1

    Of all integer pairs(x,y) that satisfy the equation 42x+55y=1, only one such pair has 100<x< 200.what is the value of x in this integer?
    This is an exercise in using Euclid's algorithm. (If you haven't come across this, Google it to see why the following calculations are being done.)

    \begin{aligned}55&=42+13\\ 42&= 3(13) + 3\\ 13&= 4(3) + 1.\end{aligned}

    Now work through those equations in the reverse order to get

    \begin{aligned}1&= 13-4(3)\\ &= 13 - 4(42-3(13)) = -4(42) + 13(13)\\ &= -4(42) + 13(55-42) = -17(42) + 13(55).\end{aligned}

    So one solution to the equation 42x+55y=1 is x=–17, y= 13. You get the other solutions by adding a multiple of 55 to x, and subtracting the same multiple of 42 from y. Thus (-17+55k)(42) + (13-42k)(55) = 1 (you can see that the multiples of k cancel, to leave you with the result from the previous displayed equation).

    Now all you have to do is to choose k so that –17+55k lies between 100 and 200.

    Quote Originally Posted by barhin View Post
    Question 2
    Let X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 194 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)
    For this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X – 12 = (multiple of 194). Then add suitable multiples of 194 to find the two smallest positive values for X.
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  3. #3
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    Hello, barhin!

    \text{(1) Of all integer pairs }(x,y)\text{ that satisfy the equation: }\: 42x+55y\:=\:1
    . . \text{only one such pair has }100<x< 200.

    . . \text{What is this value of }x\,?

    The problem can be solved without Euclid's algorithm.
    . . But, of course, this solution takes much longer.


    We have: . 42x + 55y \:=\:1

    \text{Then: }\:x \:=\:\frac{1-55y}{42} \:=\:\frac{-42y + 1 - 13y}{42} \quad\Rightarrow\quad x \:=\:-y + \frac{1-13y}{42} .[1]


    Since x is an integer, 1-13y must be a multiple of 42.

    . . 1-13y \:=\:42a \quad\Rightarrow\quad y \:=\:\frac{1-42a}{13} \quad\Rightarrow\quad y \:=\:-3a + \frac{1-3a}{13} .[2]


    Since y in an integer, 1-3a must be a multiple of 13.

    . . 1 - 3a \:=\:13b \quad\Rightarrow\quad a \:=\:\frac{1-13b}{3} \quad\Rightarrow\quad a \:=\:-4b + \frac{1-b}{3} .[3]


    Since a is an integer, 1-b must be a multiple of 3.

    . . 1-b \:=\:3k \quad\Rightarrow\quad b \:=\:1-3k


    Substitute into [3]:
    . . a \:=\:\text{-}4(1-3b) + \frac{1-(1-3k)}{3} \quad\Rightarrow\quad a \:=\:13k-4

    Substitute into [2]:
    . . y \:=\:\text{-}3(13k-4) + \frac{1-3(13k-4)}{13} \quad\Rightarrow\quad y \:=\:\text{-}42k + 13

    Substitute into [1]:
    . . x \:=\:\text{-}(\text{-}42k+13) + \frac{1-13(\text{-}42k+13)}{42} \quad\Rightarrow\quad x \:=\:55k - 17 .[4]


    Since 100 < x < 200

    . . we have:. . 100 \;<\;55k - 17 \;<\; 200

    . . . Add 17: . . . 117 \;<\;55k \;<\; 217

    Divide by 55: . . \frac{117}{55} \;<\;k \;<\;\frac{217}{55}

    . . . . . . . . . . . 2.127 \;<\; k \;<\;3.945

    . . . . . . . . . . . . . . 2 \;<\;k \;<\;4

    Hence: . k\;=\;3

    Substitute into [4]: . x \:=\:55(3) - 17 \quad\Rightarrow\quad \boxed{x \:=\:148}

    The integer pair is: . (x,y) \:=\:(148,\:\text{-}113)

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  4. #4
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    Quote Originally Posted by Opalg View Post
    This is an exercise in using Euclid's algorithm. (If you haven't come across this, Google it to see why the following calculations are being done.)

    \begin{aligned}55&=42+13\\ 42&= 3(13) + 3\\ 13&= 4(3) + 1.\end{aligned}

    Now work through those equations in the reverse order to get

    \begin{aligned}1&= 13-4(3)\\ &= 13 - 4(42-3(13)) = -4(42) + 13(13)\\ &= -4(42) + 13(55-42) = -17(42) + 13(55).\end{aligned}

    So one solution to the equation 42x+55y=1 is x=–17, y= 13. You get the other solutions by adding a multiple of 55 to x, and subtracting the same multiple of 42 from y. Thus (-17+55k)(42) + (13-42k)(55) = 1 (you can see that the multiples of k cancel, to leave you with the result from the previous displayed equation).

    Now all you have to do is to choose k so that –17+55k lies between 100 and 200.


    For this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X – 12 = (multiple of 19). Then add suitable multiples of 19 to find the two smallest positive values for X.



    Dear Soroban,
    Thank very much for useful taught on question 1 and 2 , I am how ever getting different results for the second question,will be grateful if you could throw more light or solution to the question 2 for me.It's my prayer that I will become a great mathematician like you
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  5. #5
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    Hello, barhin!

    \text{(2) Let }x_1\text{ and }x_2\text{ be the two smallest positive integers}
    \text{for which the following statement is true: }\:85x-12\text{ is a multiple of 19.}
    \text{Find }x_1+x_2.

    85x - 12 \text{ is a multiple of 19}

    \text{Hence: }\:85x-12 \:=\:19a\;\text{ for some integer }a.

    . . \text{Then: }\: a \:=\:\frac{85x-12}{19} \quad\Rightarrow\quad a \:=\:4x + \frac{9x-12}{19}

    \text{Since }a\text{ is an integer, then }9x-12\text{ must be multiple of 19.}

    \text{Hence: }\:9x-12 \:=\:19b \quad\Rightarrow\quad x \:=\:\frac{19b + 12}{9} \quad\Rightarrow\quad x \:=\:2b + 1 +\frac{b+3}{9} .[1]

    \text{Since }b\text{ is an integer, then }b+3\text{ must be a multiple of 9.}

    \text{Hence: }\:b +3 \:=\:9k \quad\Rightarrow\quad b \:=\:9k-3


    Substitute into [1]:

    . . x \;=\;2(9k-3) + 1 + \frac{(9k-3) + 3}{9} \quad\Rightarrow\quad x \:=\:19k-5


    \text{The first two positive values of }x\text{ are: }\:\begin{Bmatrix} k = 0 & \Rightarrow & x_1 \:=\:14 \\ k = 1 & \Rightarrow & x_2 \:=\:33 \end{Bmatrix}

    \text{Therefore: }\:x_1 + x_2 \:=\:14 + 33 \:=\:47

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