Weird question pertaining to HCF

Question

I encountered this question which seems weird/incomplete to me :

Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is $2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?

Can anyone please teach me concept wise how to solve it?

Answer 1

Find the prime power factorizations of the two given numbers. We get

$3240=2^3\cdot 3^4\cdot 5^1$ and

$3600=2^4\cdot 3^2\cdot 5^2$.

Let our unknown number be $n$. Because the LCM of $3240$, $3600$, and $n$ only involves the primes $2$, $3$, $5$, and $7$, we know that the prime power factorization of $n$ can involve no primes other than these.

So the only question is: how many of each?

Since the HCF of our three numbers is $36=2^2\cdot 3^2$, the highest power of $2$ that divides $n$ must be $2^2$.

The LCM has a $3^5$. Since the highest power of $3$ needed by our first two numbers is $3^4$, the highest power of $3$ that divides $n$ must be $3^5$.

Note that $5$ cannot divide $n$ since $5$ divides $3240$ and $3600$ but does not divide $36$.

Note also that since $7$ does not divide the first two numbers, the $7^2$ in the LCM must come from $n$.

It follows that $n=2^2\cdot 3^5\cdot 7^2$.

Remark: Your intuition about "not enough information" is reasonable. For example, if the HCF of the three numbers was $72$ instead of $36$, then the highest power of $2$ that divides $n$ could be $2^3$ or $2^4$, so $n$ would not be completely determined.

Answer 2

\begin{align} 3240 & = 2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot5 & & = 2^3\cdot3^4\cdot5\\ 3600 & = 2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5 & & = 2^4\cdot3^2\cdot5^2 \\ 36 & = 2\cdot2\cdot3\cdot3 & & =2^2\cdot3^2 \end{align}

The third number cannot have more than two $2$s in its prime factorization, since then the gcd would not be $36$, but at least $2$ times that. The third cannot have any $5$ in its prime factorization, since then the gcd would have a $5$ in it. The third number must have two $7$s in its prime factorization, since there's nothing else to contribute those two $7$s to the lcm. The third number must have at least two $2$s in its prime factoriation; otherwise it would not be divisible by $36$. The third number must have at least two $3$s in its prime factorization for the same reason.

So the third number could be $2\cdot2\cdot3\cdot3\cdot7\cdot7$.

It could also be $2\cdot2\cdot3\cdot3\cdot3\cdot7\cdot7$ or $2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot7\cdot7$.