Find two numbers whose sum is 20 and LCM is 24 With some guess work I found the answers to be 8 and 12.
But is there any general formula for this?
Note that the question is asked to my nephew who is at 4th grade.
 A: With an LCM of 24 one of the numbers must be 8, as $2^3$ divides 24. It can't be 16 as 16 does not divide 24.
A: You want two factors of $24$ that add up to $20$.   That means one factor must be less than or equal to $10$ (i.e. $20/2$), and the other must be between $10 $ and $20$ (inclusive) since it is equal to the first subtracted from $20$.
The only factor of $24$ between $10$ and $20$ inclusive is $12$.    $20-12 = 8$.   $8$ is a factor.
So the solution is $8$ and $12$.
A: Let $x$ and $y$ be the numbers. Note that $xy$ is the product of the LMC and the gcd, say, $d$. Then:
$$xy=24d$$
$$x+y=20$$
I don't know if at 4th grade the pupils are supposed to be able of solve systems. If it is the case, you can write
$$y=20-x$$
and then
$$x(20-x)=24d$$
Solve $x$ to get
$$x=\frac{20\pm\sqrt{400-96d}}2=10\pm2\sqrt{25-6d}$$
Therefore, $25-6d$ must be a perfect square. This only happens if $d=4$, which yields $x=8$ or $x=12$.
This is not a formula, but it makes the search much easier.
A: Let the numbers be $a,b$ and let $\gcd=d$,
then $a=pd, b=qd$
$a+b=d(p+q)=20$
also $a.b=(l.c.m)(\gcd) $
or, $pqd^2=(l.c.m)(d)$, or, $l.c.m=pqd=24$
dividing the two equations we get,
$\frac{p+q}{pq}=\frac{20}{24}=\frac{5}{6}$
or,$6(p+q)=5pq$
$5$ does not divide $6$, therefore, $5$ divides$(p+q)$, also, $p+q$ does not divide pq, or, $p+q$ divides $5$, $\implies$ $p+q=5$ and $pq =6$ and $d=4$
solve for $p$ and $q$ to get $a$ and $b$
A: With 4th grade knowledge in mind, I would say this question is about finding all the factors of $24$, because the two numbers must both be factors of $24$ or they would have a different least common multiple.
So listing the factors of $24$ we get $1,2,3,4,6,8,12,24$. Then it's easy to see which two add to $20$, although you could certainly use the exercise, if so inclined, to develop a few more rules for harder cases, and in another case you might need to be a little careful that the least common multiple is indeed still $24$.

Another approach: $\frac{20}{24} = \frac{5}{6}$ - can we write $\frac{5}{6}$ as the sum of two unit fractions? The answer is yes, $\frac{1}{2} + \frac{1}{3}$, and the numbers we seek are $\frac{24}{2}=12$ and $\frac{24}{3}=8$
A: A general algorithm
If $a+b = n$ and $\mathrm{lcm}(a,b) = c$ we let $\gcd(a,b) = d$ to get
$$ab = cd\\
n = a+b$$
Solving $b = n-a$ gives us
$$a(n-a) = cd \\
\Leftrightarrow a^2-na + cd = 0 \\
\Leftrightarrow a = \frac n2 \pm \sqrt{\frac{n^2}4 - cd}$$
So for even $n$ we must find $d$ such that $\frac{n^2}4 - cd$ is a perfect square (since $c>0$ this will amount to a finite number of possibilities). The $\pm$ is irrelevant because $b$ will take the alternate value.
If $n$ is odd, $n^2 - 4cd$ must be a perfect square instead and we obtain an analogous formula:
$$a = \frac12 (n \pm \sqrt{n^2 - 4cd})$$
In both cases, $d$ can range between $1$ and $\left\lfloor \frac{n^2}{4c} \right\rfloor$
Your case thus allows $1\le d\le \lfloor\frac{400}{96}\rfloor = 4$ and $100-24d$ must be a perfect square. $d=4$ yields $4 = 2^2$ so
$$a = 10 + \sqrt{4} = 12; \quad b = 10 - \sqrt4 = 8$$
A: Let the numbers be $a, b$.
$\gcd(a,b) = \gcd((a+b), lcm(a,b))$ (The reason is explained later)
The $lcm(a,b)$ is given. Thus, you can find the product of the numbers using the formula: $ab=lcm(a,b)*gcd(a,b)$.
$(a+b)$ is also given. Now, you can easily solve these 2 equations.
For this particular question,
$a+b=20$ ----------(1)
$gcf(20,24)=4$
$ab=24*4=96$ ----------(2)

Let the numbers be $a,b$ and their $\gcd=d$,
Then $a=pd, b=qd$, where $p,q$ are co-prime.
$a+b=d(p+q)$
$lcm(a,b)=d(pq)$
Now, because $p, q$ are co-primes, $p+q$ and $pq$ will be co-prime too. i.e., $gcd(p+q,pq)=1$. Visit this question for the explanation.
Thus, $\gcd(d(p+q),d(pq))=d$
$\implies$ $\gcd((a+b),lcm(a,b))=d=\gcd(a,b)$

I am not experienced in writing mathematics symbol markdowns. I tried my best. Please edit if required.
