# How can we find the two numbers when their sum and the sum of their HCF and LCM are given?

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?

• $48b-96h = b^2 - h^2$ 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. Aug 29, 2016 at 14:20
• @lulu The two numbers are unequal. Aug 29, 2016 at 14:24
• what do you mean with HCF? Aug 29, 2016 at 14:25
• @Dr.SonnhardGraubner Highest Common Factor (Greatest Common Divisor). Aug 29, 2016 at 14:27
• @gammatester I am looking for a concrete method of getting this solution. Aug 29, 2016 at 14:46

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

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 $l + h = 96$

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 $b = 30$.

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: $$a=\color{red}{18},\quad b=\color{red}{30}.$$
• Why is there a $d'$ in first line? Aug 30, 2016 at 12:15
• It's only a comma after the fractions. Doesn't bite :o) Aug 30, 2016 at 12:23

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 $\{a,b\}=\{a'd,\; b'd\}=\{18,30\}.$