# help (1 Viewer)

#### mathsbrain

##### Member
I'm always getting stuck with questions like these, not in standard textbooks i don't think, how can i improve on these, like any books out there or website?(and also if you can shed light on how to solve it plz...)
Q: Determine all possible pairs of positive integers such that their sum is 2623 and the quotient of the lowest common multiple and highest common factor is 280 with remainder zero.

#### mathsbrain

##### Member
I'm always getting stuck with questions like these, not in standard textbooks i don't think, how can i improve on these, like any books out there or website?(and also if you can shed light on how to solve it plz...)
Q: Determine all possible pairs of positive integers such that their sum is 2623 and the quotient of the lowest common multiple and highest common factor is 280 with remainder zero.
and another similar one...
Q: positive integers x and y satisfy the equation 3x+5y=2xy-1, find all possible values of x-y.

#### icycledough

##### Well-Known Member
It's interesting because these types of questions aren't standard to be assessed in any level of Maths (Standard, Advanced, 3U, 4U) in Year 12, let alone Year 11. If anything, it would fall under a 'Miscellaneous' category or a question similar to those found in AMC or ICAS. The problem isn't the difficulty I'd say, but rather, how to approach it from a unique viewpoint

#### cossine

##### Active Member
I'm always getting stuck with questions like these, not in standard textbooks i don't think, how can i improve on these, like any books out there or website?(and also if you can shed light on how to solve it plz...)
Q: Determine all possible pairs of positive integers such that their sum is 2623 and the quotient of the lowest common multiple and highest common factor is 280 with remainder zero.
Real maths and competition maths are different.

#### mathsbrain

##### Member
It's interesting because these types of questions aren't standard to be assessed in any level of Maths (Standard, Advanced, 3U, 4U) in Year 12, let alone Year 11. If anything, it would fall under a 'Miscellaneous' category or a question similar to those found in AMC or ICAS. The problem isn't the difficulty I'd say, but rather, how to approach it from a unique viewpoint
right, so could you shed some light on how to solve the above questions?

#### icycledough

##### Well-Known Member
right, so could you shed some light on how to solve the above questions?
With the second question you posted, the one where you have to find the values of x - y, I just tried getting an equation for y in terms of x. But I'm not too sure if that's the right way to go about it, because you could be testing thousands of possible combinations. I'm sure there's a different way to go about it which I just haven't thought about yet.

can anyone help?

#### mathsbrain

##### Member
hmm seems this question is slightly harder though coz it cant be factorised?

#### CM_Tutor

##### Moderator
Moderator
and another similar one...
Q: positive integers x and y satisfy the equation 3x+5y=2xy-1, find all possible values of x-y.
I don't know what approach they seek, but my approach (potentially going beyond year 9 or 10 Maths*) is as follows:

Rearranging the given equation to make $\bg_white y$ the subject, I get that

$\bg_white y = \cfrac{3x+1}{2x-5}$

and sketching this gives a rectangular hyperbola with asymptotes at $\bg_white x = \cfrac{5}{2}$ and $\bg_white y=\cfrac{3}{2}$.

From this graph, a few things are immediately clear:
• as $\bg_white x$ increases, $\bg_white y$ decreases
• we can only have both $\bg_white x$ and $\bg_white y$ for $\bg_white x > \cfrac{5}{2}$ and hence the minimum possible value for $\bg_white x$ is 3.
• as $\bg_white x$ increases, $\bg_white y$ approaches the value 1.5, and so I can find an upper bound for $\bg_white x$ by examining the case $\bg_white y = 2$:
\bg_white \begin{align*} y = \cfrac{3x+1}{2x - 5} &= 2 \\ 3x+1 &= 2(2x - 5) \qquad \text{provided x \ne \cfrac{5}{2}, so we don't multiply by 0} \\ 3x + 1 &= 4x - 10 \\ 1 + 10 &= 4x - 3x \\ x &= 11 \end{align*}

• So, if $\bg_white x > 11$ then $\bg_white y < 2$ (as $\bg_white y$ decreases when $\bg_white x$ increases), and since $\bg_white y$ is bounded below by 1.5 for positive $\bg_white x$, $\bg_white y$ cannot be an integer if $\bg_white x > 11$. In other words, all solutions must satisfy $\bg_white 3 \leqslant x \leqslant 11$.
• We already have one solution, $\bg_white y = 2$ and $\bg_white x = 11$, for which $\bg_white x - y = 9$
We could now find all solutions by trial-and-error by sequentially testing $\bg_white x = 3, 4, 5, ..., 11$. This gives us that $\bg_white 3x+1 = 10, 13, 16, ..., 34$ and that $\bg_white 2x-5 = 1, 3, 5, ..., 17$, and thus

$\bg_white y = \color{red}\cfrac{10}{1} = 10\color{black},\ \cfrac{13}{3},\ \cfrac{16}{5},\ \cfrac{19}{7},\ \cfrac{22}{9},\ \cfrac{25}{11},\ \cfrac{28}{13},\ \cfrac{31}{15},\ \text{or}\ \color{red} \cfrac{34}{17} = 2\color{black}$

So, there are two solutions:

\bg_white \begin{align*} x = 3 \quad \text{and} \quad y = 10 \qquad &\implies \qquad x - y = 3 - 10 = -7 \\ x = 11 \quad \text{and} \quad y = 2 \qquad &\implies \qquad x - y = 11 - 2 = 9 \end{align*}

If the requirement that $\bg_white x$ and $\bg_white y$ be positive is removed, but they are still required to be integers, two other solutions appear:

\bg_white \begin{align*} x = 2 \quad \text{and} \quad y = -7 \qquad &\implies \qquad x - y = 2 - -7 = 9 \\ x = -6 \quad \text{and} \quad y = 1 \qquad &\implies \qquad x - y = -6 - 1 = -7 \end{align*}

but it does not add any extra possible values for $\bg_white x - y$, which can only be -7 or 9.

* A note on "potentially going beyond year 9 or 10 Maths"... I originally defined a function $\bg_white f(x) = x - y= x - \cfrac{3x+1}{2x-5}$ which has an oblique asymptote of $\bg_white y=x-\cfrac{3}{2}$ and then used the fact that $\bg_white f(x)$ must be an integer to constrain the possible domain of $\bg_white x$... once $\bg_white f(x)$ is sufficiently close to the asymptote, any integer value of $\bg_white x$ will give a non-integer value of $\bg_white f(x)$. As I was typing this up, however, I realised that I could simplify the method by looking at just the above hyperbola, which makes the level of the Maths involved suited to junior high school so long as hyperbolas have been covered.

Last edited:

#### mathsbrain

##### Member
I don't know what approach they seek, but my approach (potentially going beyond year 9 or 10 Maths*) is as follows:

Rearranging the given equation to make $\bg_white y$ the subject, I get that

$\bg_white y = \cfrac{3x+1}{2x-5}$

and sketching this gives a rectangular hyperbola with asymptotes at $\bg_white x = \cfrac{5}{2}$ and $\bg_white y=\cfrac{3}{2}$.

View attachment 32730

From this graph, a few things are immediately clear:
• as $\bg_white x$ increases, $\bg_white y$ decreases
• we can only have both $\bg_white x$ and $\bg_white y$ for $\bg_white x > \cfrac{5}{2}$ and hence the minimum possible value for $\bg_white x$ is 3.
• as $\bg_white x$ increases, $\bg_white y$ approaches the value 1.5, and so I can find an upper bound for $\bg_white x$ by examining the case $\bg_white y = 2$:
\bg_white \begin{align*} y = \cfrac{3x+1}{2x - 5} &= 2 \\ 3x+1 &= 2(2x - 5) \qquad \text{provided x \ne \cfrac{5}{2}, so we don't multiply by 0} \\ 3x + 1 &= 4x - 10 \\ 1 + 10 &= 4x - 3x \\ x &= 11 \end{align*}

• So, if $\bg_white x > 11$ then $\bg_white y < 2$ (as $\bg_white y$ decreases when $\bg_white x$ increases), and since $\bg_white y$ is bounded below by 1.5 for positive $\bg_white x$, $\bg_white y$ cannot be an integer if $\bg_white x > 11$. In other words, all solutions must satisfy $\bg_white 3 \leqslant x \leqslant 11$.
• We already have one solution, $\bg_white y = 2$ and $\bg_white x = 11$, for which $\bg_white x - y = 9$
We could now find all solutions by trial-and-error by sequentially testing $\bg_white x = 3, 4, 5, ..., 11$. This gives us that $\bg_white 3x+1 = 10, 13, 16, ..., 34$ and that $\bg_white 2x-5 = 1, 3, 5, ..., 17$, and thus

$\bg_white y = \color{red}\cfrac{10}{1} = 10\color{black},\ \cfrac{13}{3},\ \cfrac{16}{5},\ \cfrac{19}{7},\ \cfrac{22}{9},\ \cfrac{25}{11},\ \cfrac{28}{13},\ \cfrac{31}{15},\ \text{or}\ \color{red} \cfrac{34}{17} = 2\color{black}$

So, there are two solutions:

\bg_white \begin{align*} x = 3 \quad \text{and} \quad y = 10 \qquad &\implies \qquad x - y = 3 - 10 = -7 \\ x = 11 \quad \text{and} \quad y = 2 \qquad &\implies \qquad x - y = 11 - 2 = 9 \end{align*}

If the requirement that $\bg_white x$ and $\bg_white y$ be positive is removed, but they are still required to be integers, two other solutions appear:

\bg_white \begin{align*} x = 2 \quad \text{and} \quad y = -7 \qquad &\implies \qquad x - y = 2 - -7 = 9 \\ x = -6 \quad \text{and} \quad y = 1 \qquad &\implies \qquad x - y = -6 - 1 = -7 \end{align*}

but it does not add any extra possible values for $\bg_white x - y$, which can only be -7 or 9.

* A note on "potentially going beyond year 9 or 10 Maths"... I originally defined a function $\bg_white f(x) = x - y= x - \cfrac{3x+1}{2x-5}$ which has an oblique asymptote of $\bg_white y=x-\cfrac{3}{2}$ and then used the fact that $\bg_white f(x)$ must be an integer to constrain the possible domain of $\bg_white x$... once $\bg_white f(x)$ is sufficiently close to the asymptote, any integer value of $\bg_white x$ will give a non-integer value of $\bg_white f(x)$. As I was typing this up, however, I realised that I could simplify the method by looking at just the above hyperbola, which makes the level of the Maths involved suited to junior high school so long as hyperbolas have been covered.
I don't know what approach they seek, but my approach (potentially going beyond year 9 or 10 Maths*) is as follows:

Rearranging the given equation to make $\bg_white y$ the subject, I get that

$\bg_white y = \cfrac{3x+1}{2x-5}$

and sketching this gives a rectangular hyperbola with asymptotes at $\bg_white x = \cfrac{5}{2}$ and $\bg_white y=\cfrac{3}{2}$.

View attachment 32730

From this graph, a few things are immediately clear:
• as $\bg_white x$ increases, $\bg_white y$ decreases
• we can only have both $\bg_white x$ and $\bg_white y$ for $\bg_white x > \cfrac{5}{2}$ and hence the minimum possible value for $\bg_white x$ is 3.
• as $\bg_white x$ increases, $\bg_white y$ approaches the value 1.5, and so I can find an upper bound for $\bg_white x$ by examining the case $\bg_white y = 2$:
\bg_white \begin{align*} y = \cfrac{3x+1}{2x - 5} &= 2 \\ 3x+1 &= 2(2x - 5) \qquad \text{provided x \ne \cfrac{5}{2}, so we don't multiply by 0} \\ 3x + 1 &= 4x - 10 \\ 1 + 10 &= 4x - 3x \\ x &= 11 \end{align*}

• So, if $\bg_white x > 11$ then $\bg_white y < 2$ (as $\bg_white y$ decreases when $\bg_white x$ increases), and since $\bg_white y$ is bounded below by 1.5 for positive $\bg_white x$, $\bg_white y$ cannot be an integer if $\bg_white x > 11$. In other words, all solutions must satisfy $\bg_white 3 \leqslant x \leqslant 11$.
• We already have one solution, $\bg_white y = 2$ and $\bg_white x = 11$, for which $\bg_white x - y = 9$
We could now find all solutions by trial-and-error by sequentially testing $\bg_white x = 3, 4, 5, ..., 11$. This gives us that $\bg_white 3x+1 = 10, 13, 16, ..., 34$ and that $\bg_white 2x-5 = 1, 3, 5, ..., 17$, and thus

$\bg_white y = \color{red}\cfrac{10}{1} = 10\color{black},\ \cfrac{13}{3},\ \cfrac{16}{5},\ \cfrac{19}{7},\ \cfrac{22}{9},\ \cfrac{25}{11},\ \cfrac{28}{13},\ \cfrac{31}{15},\ \text{or}\ \color{red} \cfrac{34}{17} = 2\color{black}$

So, there are two solutions:

\bg_white \begin{align*} x = 3 \quad \text{and} \quad y = 10 \qquad &\implies \qquad x - y = 3 - 10 = -7 \\ x = 11 \quad \text{and} \quad y = 2 \qquad &\implies \qquad x - y = 11 - 2 = 9 \end{align*}

If the requirement that $\bg_white x$ and $\bg_white y$ be positive is removed, but they are still required to be integers, two other solutions appear:

\bg_white \begin{align*} x = 2 \quad \text{and} \quad y = -7 \qquad &\implies \qquad x - y = 2 - -7 = 9 \\ x = -6 \quad \text{and} \quad y = 1 \qquad &\implies \qquad x - y = -6 - 1 = -7 \end{align*}

but it does not add any extra possible values for $\bg_white x - y$, which can only be -7 or 9.

* A note on "potentially going beyond year 9 or 10 Maths"... I originally defined a function $\bg_white f(x) = x - y= x - \cfrac{3x+1}{2x-5}$ which has an oblique asymptote of $\bg_white y=x-\cfrac{3}{2}$ and then used the fact that $\bg_white f(x)$ must be an integer to constrain the possible domain of $\bg_white x$... once $\bg_white f(x)$ is sufficiently close to the asymptote, any integer value of $\bg_white x$ will give a non-integer value of $\bg_white f(x)$. As I was typing this up, however, I realised that I could simplify the method by looking at just the above hyperbola, which makes the level of the Maths involved suited to junior high school so long as hyperbolas have been covered.
thanks so much! cant believe they ask this in year 8 SCHOOL EXAM at my school! Are you able to help with the first question plz?

#### CM_Tutor

##### Moderator
Moderator
Are you able to help with the first question plz?
You need to notice that the prime factorisation of 2623 is 1 x 43 x 61

Let $\bg_white Z$ be the lowest common multiple of $\bg_white x$ and $\bg_white y$, which means that $\bg_white Z = ax$ and $\bg_white Z = by$, for some integers $\bg_white a$ and $\bg_white b$. It follows that:

\bg_white \begin{align*} x + y &= \cfrac{Z}{a} + \cfrac{Z}{b} \\ &=\cfrac{Z(a + b)}{ab} \\ \text{And, since x + y = 2623:} \qquad \cfrac{Z(a + b)}{ab} &= 2623 \times 1 = 43 \times 61\end{align*}
We thus have two expressions that multiply to 2623. They can't be 1 and 2623 because we need LCM = 280 x HCF, so we are left with:

\bg_white \begin{align*} \textbf{Option 1a} \qquad \qquad \cfrac{Z}{a} = 43 \qquad &\text{and} \qquad \cfrac{a + b}{b} = 61 \\ \\ \textbf{Option 1b} \qquad \qquad \cfrac{Z}{a} = 61 \qquad &\text{and} \qquad \cfrac{a + b}{b} = 43 \\ \\ \textbf{Option 2a} \qquad \qquad \cfrac{Z}{b} = 43 \qquad &\text{and} \qquad \cfrac{a + b}{a} = 61 \\ \\ \textbf{Option 2b} \qquad \qquad \cfrac{Z}{b} = 61 \qquad &\text{and} \qquad \cfrac{a + b}{a} = 43 \\ \\ \textbf{Option 3a} \qquad \qquad \cfrac{Z}{ab} = 43 \qquad &\text{and} \qquad a + b = 61 \\ \\ \textbf{Option 3b} \qquad \qquad \cfrac{Z}{ab} = 61 \qquad &\text{and} \qquad a + b = 43 \end{align*}

If you look carefully, you will find that Options 2a and 2b are duplicates of 1a and 1b, in that they swap the results for $\bg_white x$ and $\bg_white y$, but which is which was already arbitrary. The way to assess the solutions from the options is whether they produce the required ratio of LCM to HCF.

Option 1a

$\bg_white Z = 43a = ax \quad \implies \quad x = 43 \quad \implies \quad y = 2623 - 43 = 2580$

$\bg_white 2580 = 60 \times 43 \quad \implies \quad \text{HCF} = 43\ \text{and LCM} = 60 \times 43$

$\bg_white \text{Thus,} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{60 \times 43}{43} = 60 \ne 280 \qquad \textbf{NOT a valid solution}$

Option 1b

$\bg_white Z = 61a = ax \quad \implies \quad x = 61 \quad \implies \quad y = 2623 - 61 = 2562$

$\bg_white 2562 = 42 \times 61 \quad \implies \quad \text{HCF} = 61\ \text{and LCM} = 42 \times 61$

$\bg_white \text{Thus,} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{42 \times 61}{61} = 42 \ne 280 \qquad \textbf{NOT a valid solution}$

Option 3a

$\bg_white Z = 43ab = ax \quad \implies \quad x = 43b \quad \implies \quad y = 2623 - 43b = 43(61 - b)$

$\bg_white \text{The HCF of 43b and 43(61 - b) is 43 unless b and 61 - b have a common factor (which they don't for any integer 1 < b < 60).}$

$\bg_white \text{The LCM of 43b and 43(61 - b) is 43b(61 - b) unless b and 61 - b have a common factor.}$

\bg_white \begin{align*} \text{Thus, we need} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{43b(61 - b)}{43} &= 280 \\ 61b - b^2 &= 280 \\ b^2 - 61b + 280 &= 0 \\ (b - 56)(b - 5) &= 0 \end{align*}

\bg_white \begin{align*} \text{Our possible solutions are} \quad &b = 5 \quad \implies \quad x = 43 \times 5 = 215\ \text{and}\ y = 43 \times 56 = 2408 \qquad \text{with LCM = 12,040 and HCF = 43} \\ &b = 56 \quad \implies \quad x = 43 \times 56 = 2408\ \text{and}\ y = 43 \times 5 = 215 \qquad \text{which is identical to the first solution} \end{align*}

Option 3b

$\bg_white Z = 61ab = ax \quad \implies \quad x = 61b \quad \implies \quad y = 2623 - 61b = 61(43 - b)$

$\bg_white \text{The HCF of 61b and 61(43 - b) is 61 unless b and 43 - b have a common factor (which they don't for any integer 1 < b < 42).}$

$\bg_white \text{The LCM of 61b and 61(43 - b) is 61b(43 - b) unless b and 61 - b have a common factor.}$

\bg_white \begin{align*} \text{Thus, we need} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{61b(43 - b)}{61} &= 280 \\ 43b - b^2 &= 280 \\ b^2 - 43b + 280 &= 0 \\ (b - 35)(b - 8) &= 0 \end{align*}

\bg_white \begin{align*} \text{Our possible solutions are} \quad &b = 8 \quad \implies \quad x = 61 \times 8 = 488\ \text{and}\ y = 61 \times 35 = 2135 \qquad \text{with LCM = 17,080 and HCF = 61} \\ &b = 35 \quad \implies \quad x = 61 \times 35 = 2135\ \text{and}\ y = 61 \times 8 = 488 \qquad \text{which is identical to the first solution} \end{align*}

Overall

There are two solutions to this problem, the first solution being the pair of numbers 215 and 2408, and the second solution being the pair of numbers 488 and 2135.

#### mathsbrain

##### Member
You need to notice that the prime factorisation of 2623 is 1 x 43 x 61

Let $\bg_white Z$ be the lowest common multiple of $\bg_white x$ and $\bg_white y$, which means that $\bg_white Z = ax$ and $\bg_white Z = by$, for some integers $\bg_white a$ and $\bg_white b$. It follows that:

\bg_white \begin{align*} x + y &= \cfrac{Z}{a} + \cfrac{Z}{b} \\ &=\cfrac{Z(a + b)}{ab} \\ \text{And, since x + y = 2623:} \qquad \cfrac{Z(a + b)}{ab} &= 2623 \times 1 = 43 \times 61\end{align*}
We thus have two expressions that multiply to 2623. They can't be 1 and 2623 because we need LCM = 280 x HCF, so we are left with:

\bg_white \begin{align*} \textbf{Option 1a} \qquad \qquad \cfrac{Z}{a} = 43 \qquad &\text{and} \qquad \cfrac{a + b}{b} = 61 \\ \\ \textbf{Option 1b} \qquad \qquad \cfrac{Z}{a} = 61 \qquad &\text{and} \qquad \cfrac{a + b}{b} = 43 \\ \\ \textbf{Option 2a} \qquad \qquad \cfrac{Z}{b} = 43 \qquad &\text{and} \qquad \cfrac{a + b}{a} = 61 \\ \\ \textbf{Option 2b} \qquad \qquad \cfrac{Z}{b} = 61 \qquad &\text{and} \qquad \cfrac{a + b}{a} = 43 \\ \\ \textbf{Option 3a} \qquad \qquad \cfrac{Z}{ab} = 43 \qquad &\text{and} \qquad a + b = 61 \\ \\ \textbf{Option 3b} \qquad \qquad \cfrac{Z}{ab} = 61 \qquad &\text{and} \qquad a + b = 43 \end{align*}

If you look carefully, you will find that Options 2a and 2b are duplicates of 1a and 1b, in that they swap the results for $\bg_white x$ and $\bg_white y$, but which is which was already arbitrary. The way to assess the solutions from the options is whether they produce the required ratio of LCM to HCF.

Option 1a

$\bg_white Z = 43a = ax \quad \implies \quad x = 43 \quad \implies \quad y = 2623 - 43 = 2580$

$\bg_white 2580 = 60 \times 43 \quad \implies \quad \text{HCF} = 43\ \text{and LCM} = 60 \times 43$

$\bg_white \text{Thus,} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{60 \times 43}{43} = 60 \ne 280 \qquad \textbf{NOT a valid solution}$

Option 1b

$\bg_white Z = 61a = ax \quad \implies \quad x = 61 \quad \implies \quad y = 2623 - 61 = 2562$

$\bg_white 2562 = 42 \times 61 \quad \implies \quad \text{HCF} = 61\ \text{and LCM} = 42 \times 61$

$\bg_white \text{Thus,} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{42 \times 61}{61} = 42 \ne 280 \qquad \textbf{NOT a valid solution}$

Option 3a

$\bg_white Z = 43ab = ax \quad \implies \quad x = 43b \quad \implies \quad y = 2623 - 43b = 43(61 - b)$

$\bg_white \text{The HCF of 43b and 43(61 - b) is 43 unless b and 61 - b have a common factor (which they don't for any integer 1 < b < 60).}$

$\bg_white \text{The LCM of 43b and 43(61 - b) is 43b(61 - b) unless b and 61 - b have a common factor.}$

\bg_white \begin{align*} \text{Thus, we need} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{43b(61 - b)}{43} &= 280 \\ 61b - b^2 &= 280 \\ b^2 - 61b + 280 &= 0 \\ (b - 56)(b - 5) &= 0 \end{align*}

\bg_white \begin{align*} \text{Our possible solutions are} \quad &b = 5 \quad \implies \quad x = 43 \times 5 = 215\ \text{and}\ y = 43 \times 56 = 2408 \qquad \text{with LCM = 12,040 and HCF = 43} \\ &b = 56 \quad \implies \quad x = 43 \times 56 = 2408\ \text{and}\ y = 43 \times 5 = 215 \qquad \text{which is identical to the first solution} \end{align*}

Option 3b

$\bg_white Z = 61ab = ax \quad \implies \quad x = 61b \quad \implies \quad y = 2623 - 61b = 61(43 - b)$

$\bg_white \text{The HCF of 61b and 61(43 - b) is 61 unless b and 43 - b have a common factor (which they don't for any integer 1 < b < 42).}$

$\bg_white \text{The LCM of 61b and 61(43 - b) is 61b(43 - b) unless b and 61 - b have a common factor.}$

\bg_white \begin{align*} \text{Thus, we need} \quad \cfrac{\text{LCM}}{\text{HCF}} = \cfrac{61b(43 - b)}{61} &= 280 \\ 43b - b^2 &= 280 \\ b^2 - 43b + 280 &= 0 \\ (b - 35)(b - 8) &= 0 \end{align*}

\bg_white \begin{align*} \text{Our possible solutions are} \quad &b = 8 \quad \implies \quad x = 61 \times 8 = 488\ \text{and}\ y = 61 \times 35 = 2135 \qquad \text{with LCM = 17,080 and HCF = 61} \\ &b = 35 \quad \implies \quad x = 61 \times 35 = 2135\ \text{and}\ y = 61 \times 8 = 488 \qquad \text{which is identical to the first solution} \end{align*}

Overall

There are two solutions to this problem, the first solution being the pair of numbers 215 and 2408, and the second solution being the pair of numbers 488 and 2135.
wow this is a genius at work! and your from year 12?

#### idkkdi

##### Well-Known Member
wow this is a genius at work! and your from year 12?
he finished a phd lol

#### CM_Tutor

##### Moderator
Moderator
wow this is a genius at work! and your from year 12?
I did indeed complete year 12, including MX2 and chemistry in my studies, but I've done some more study since then, as @idkkdi has noted.

The first answer took me some time, and the second even longer to work through. Neither is a remotely sensible test of understanding at year 8 level, though that's not to say that there can't be a reason to ask them. There was a question on a maths exam that I sat in year 9 that I couldn't do then and I am pretty sure I couldn't do now!