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So, the question is that the sum of HCF and LCM is $96$ and the sum of the numbers is $48$. We need to find the numbers.

Here is my attempt to this question:

Let the numbers be $a$ and $b$ and their LCM and HCF be $l$ and $h$ respectively. So, the 3 equations that we have are

$$a + b = 48$$ $$l + h = 96$$ $$ab = lh$$

Substituting the value of $a$ as $48 - b$ and $l$ as $96 - h$ in $ab = lh$ we get $$(48 - b)b = (96 - h)h \\ \implies 48b - b^2 = 96h - h^2 \\ \implies 48b - 96h = b^2 - h^2$$

On comparing LHS with RHS we get $b$ as $48$ and $h$ as $96$. However, this would mean that LCM and $a$ are $0$ which is not true as LCM can't be less the HCF or the numbers. Is there some other way of doing it?

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  • $\begingroup$ $48b-96h = b^2 - h^2$ is not an equality of polynomials, so you cannot compare "coefficients" like that. It's an equation, which means that we have to find some $b$ and $h$ that makes it work. You've found one pair, but hopefully there are others. $\endgroup$ – Arthur Aug 29 '16 at 14:20
  • $\begingroup$ @lulu The two numbers are unequal. $\endgroup$ – Parth Aug 29 '16 at 14:24
  • $\begingroup$ what do you mean with HCF? $\endgroup$ – Dr. Sonnhard Graubner Aug 29 '16 at 14:25
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    $\begingroup$ @Dr.SonnhardGraubner Highest Common Factor (Greatest Common Divisor). $\endgroup$ – Parth Aug 29 '16 at 14:27
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    $\begingroup$ @gammatester I am looking for a concrete method of getting this solution. $\endgroup$ – Parth Aug 29 '16 at 14:46
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$$a + b = 48$$ $$l + h = 96$$ $$ab = lh$$

So $$l = ab/h$$

Let $a = ph$, $b = qh$

Therefore $$ph + qh = 48$$ $$pqh + h = 96$$ Thus $$pqh + h = 2(ph + qh)$$ $$pq + 1 = 2p + 2q$$ $$pq -2p -2q = -1$$ By adding 4 to both sides we can factor the LHS $$pq -2p -2q + 4 = 3$$ $$(p-2)(q-2) = 3$$ 3 is prime, so one of $p-2$ and $q-2$ must be 3 and the other must be 1 (since we're working with positive integers).

WLOG, let $p-2 = 3$ and $q-2 = 1$ Therefore $p=5$ and $q=3$

$p+q=8$ and since $(p + q)h = 48$

$h = 48/8 = 6$

So $a = 5.6 = 30$ and $b = 3.6 = 18$

and $l = LCM(5.6, 3.6) = 5.18 = 30.3 = 90$

Hence $l + h = 96$

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So, you have $$b^2 - 48b + (96h - h^2) = 0$$ Using the quadratic formula to get $b$ in terms of $h$, $$b = 24 \pm \sqrt{h^2 - 96h + 576} = 24 \pm \sqrt{(h-90)(h-6)+36}$$ We want what's under the root sign to be a square, and also we need $h \leq 48$. Using the expression on the right, $h = 6$ does the job, giving us $b = 18$ or $b = 30$.

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Set $d=\gcd(a,b)$, $\;a'=\dfrac ad$, $\;b'=\dfrac bd$, $\;m=\operatorname{lcm}(a,b)$. We can translate the hypotheses as $$(a'+b')d=48,\qquad m+d=(a'b'+1)d=96.$$ We deduce $a'b'+1=2(a'+b')$. Observe this implies $a'b'$ is odd, hence $a'$ and $b'$ are. Further, $a $ and $b $ can't be equal, since this leads to $m=d=a=b=24$, and in this case $m+d\neq 96$.

So $a'+b'\ge 4$, and $d$ is a divisor of $48,{}\le 12$. We'll determine the values of $a'+b'$ and $a'b'$ for all possible values of $d$. The complete list of such divisors is

  • $1,2, 4, 8$ (not divisible by $3$). Here are the corresponding values of $a'+b'$ and $a'b'$, and the quadratic equation $a'$ and $b'$ are roots of: $$\begin{array}{cccl} d &a'+b'&a'b'&a',b'\text{ sol. of}\\ \hline 1&48&95&t^2-48t+95\\ 2&24&47 &t^2-24t+47\\ 4&12&23&t^2-12t+23 \\ 8&6&11&t^2-6t+11 \end{array}$$ For $d=1$, the discriminant is equal to $481$, which is not a perfect square. For the other cases, the product of the roots is a prime number, hence integer roots should be $a'=1$ and $b'=47,23$ or $11$ respectively. However $1$ is not a root of any of these equations.
  • $3,6,12$ (divisible by $3$). Let's draw the same table as above: $$\begin{array}{cccl} d &a'+b'&a'b'&a',b'\text{ sol. of}\\ \hline 3&16&31&t^2-16t+31\\ 6&8&15 &t^2-8t+15\\ 12&4&7&t^2-4t+7 \end{array}$$ The first and third equations have no integer solution, since $1$ is not a root thereof. The second equation has roots $3$ and $5$. Whence the unique solution: $$a=\color{red}{18},\quad b=\color{red}{30}.$$
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  • $\begingroup$ Why is there a $d'$ in first line? $\endgroup$ – Parth Aug 30 '16 at 12:15
  • $\begingroup$ It's only a comma after the fractions. Doesn't bite :o) $\endgroup$ – Bernard Aug 30 '16 at 12:23
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Let $d=\gcd (a,b).$ Let $e=$lcm$(a,b).$ We have $a=a'd$ and $b=b'd$ where $a',b'$ are integers. Since lcm$(a,b)\cdot \gcd (a,b)=|ab|,$ we have $de=|a'b'd^2|.$ Since $d> 0,$ we have $e=|a'b'|d.$

So $d(|a'b'|+1)=|a'b'|d+d=e+d=96=2(48)=2(a+ b)=2(a'd+b'd)=2d(a'+b').$

Since $d\ne 0,$ we have $|a'b'|+1=2a'+2b',$ equivalently $(\;a'b'>0\land (a'-2)(b'-2)=3\;)\lor (\;a'b'<0\land (a'+2)(b'+2)=3\;).$

Therefore $$\bullet \quad (\; a'b'>0\land \{a',b'\}\in \{\{3,5\}, \{1,-1\}\;)\lor (\;a'b'<0 \land \{a',b'\}\in \{\{1,-1\},\{-5,-3\}\}.$$ Which reduces to $\{a'b'\}\in \{\{3,5\},\{1,-1\}\}.$ The second possibility $\{a',b'\}=\{1,-1\}$ is rejected because it implies $48=a+b=d(a'+b')=0.$

Therefore $48=d(a'+b')=d(3+5).$ So $d=6,$ and $\{a,b\}=\{a'd,\; b'd\}=\{18,30\}.$

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