# Math Help - Integers

1. ## 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)

2. Originally Posted by barhin
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.

Originally Posted by barhin
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.

3. 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

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

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: . $(x,y) \:=\:(148,\:\text{-}113)$

4. Originally Posted by Opalg
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

5. 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]:

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

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

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