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If L.C.M of three natural numbers $a, b, c$ is $p^2q^2r^2$(where $p, q, r$ are different prime numbers) such that sum of all possible triplets (a,b,c) are given by $m^n$ (where $m$ and $n$ are prime), then find the value of ($m-4n$)

This is how solved.
As the L.C.M of the given numbers is $p^2q^2r^2$, this means that none of the numbers a,b,c can have $p^0 orp^1,q^0orq^1,r^0 or r^1$ as their factors.
But they can have $p,q,r$ as their factors in powers of two or higher.
For example if $a=p^2q^5r^7$,$b=p^7q^2r^3$ and $c=p^9q^8r^2$ then the L.C.M of a,b,c is $p^2q^2r^2$.
Hence there should be infinite solution i.e. infinite ordered triplets.
But the author has provided with the following solution:
L.C.M. $p^2$ can be obtained from the numbers a, b, c in the following combinations. $$p^2 p^2 p^2 ⇒ 1 arrangement$$
$$p^2 p^1 p^2 ⇒ 3 arrangement$$ $$p^2 p^1 p^1 ⇒ 1 arrangement$$ $$p^2 p^2 p^0 ⇒ 3 arrangement$$ $$p^2 p^1 p^0 ⇒ 6 arrangement$$ $$p^2 p^0 p^0 ⇒ 3 arrangement$$ $$———————————––— 19 arrangement$$

So if L.C.M. is $p^2q^2r^2$ total combinations are = $19^3$ = $m^n$ ⇒ m – 4n = 7.
This is where I am confused and unable to comprehend as why the solution is provided in such a way, because in the above arrangements L.C.M will not be $p^2q^2r^2$.

Bill Dubuque
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sameed hussain
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    In your argument as to how to solve the problem, it seems you are using the $\gcd$ instead of the $\operatorname{lcm}$. – John Omielan May 23 '21 at 02:40
  • @ John Omielan, an example in the edited question can show you why I thought like that. – sameed hussain May 23 '21 at 02:54
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    Thanks for including an example. One minor point is that the problem says "... such that sum of all possible triplets (a,b,c) ...", where "sum" really means something like "number" instead. Also, in the author's statements about the number of combinations, in $$p^2 p^1 p^1 ⇒ 1 arrangement$$ it should be $3$ arrangements instead. With this change, adding up the number of arrangements will then given you a total of $19$. – John Omielan May 23 '21 at 03:14
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    @John Omielan. Thanks for pointing out the error. – sameed hussain May 23 '21 at 03:18

2 Answers2

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The LCM of a set of numbers is given by the product of common prime factors in their highest power. So the LCM being $p^2q^2r^2$ implies that 2 is the highest power of each of $p$,$q$ and $r$ in the prime factorizations of $a$,$b$ and $c$. So powers greater than 2 aren't considered.

Hope it helps!

Gaurav Chandan
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With your example of $a=p^2q^5r^7$, $b=p^7q^2r^3$ and $c=p^9q^8r^2$, then $\operatorname{lcm}(a,b,c) = p^9 q^8 r^7$. Instead, your stated result is actually $\gcd(a,b,c) = p^2 q^2 r^2$. Remember that $\operatorname{lcm}$ is the least common multiple, i.e., the smallest positive integer which is a multiple of $a$, $b$ and $c$. The value you got of $p^2 q^2 r^2$ is the greatest common divisor, i.e., the largest positive integer which is a divisor of $a$, $b$ and $c$.

With the prime factors decomposition, the exponents of each of the primes in $\gcd(a,b,c)$ is the minimum among those of $a$, $b$ and $c$, while the exponents of each of the primes in $\operatorname{lcm}(a,b,c)$ is the maximum among those of $a$, $b$ and $c$.

John Omielan
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