# The G.C.D of two numbers is $23$. The other two factors of the L.C.M are $13$ and $14$.

The G.C.D of two numbers is $23$. $13$ and $14$ are also factors of the L.C.M [out of some unknown number of factors]. What is the larger number?

The thing I am able to infer is that both numbers are divisible by $23$. Then this means that their L.C.M is divisible by all $23$, $13$ and $14$. So, may I assume that the L.C.M is just $23 \cdot 13 \cdot 14$? And if so, how?

Please show me how to go about the rest of the question.

UPDATE $\ \ \ \$I found some assistance from a friend outside M.SE and his answer was correct: The two numbers are $23 \cdot 13$ and $23 \cdot 14$.

This question is from the previous year's papers of a competitive exam. Also, if possible, please retag the question accordingly.

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I find the statement "The other two factors of the LCM are 13 and 14" somewhat confusing, as 14 is not prime. Why does it not say "The other factor is 182" or "The other prime factors are 13, 7, and 2"? I'm concerned that there might have been an error in transcription. – Chris Pressey Dec 12 '12 at 14:29
The other two factors of the lcm? How can $2,7,26,46$ not be factors of the lcm? And if you mean prime factors, $14$ is never a prime factor. – Marc van Leeuwen Dec 12 '12 at 14:31
Okay, I'd edit the question. – Πάρτη Κοηλί Dec 12 '12 at 14:35

$\def\lcm{\operatorname{lcm}}$Let $a,b \in \mathbb N$ the two numbers. We have $\lcm(a,b) =2\cdot 7 \cdot 13\cdot 23$, $\gcd(a,b) = 23$. Hence both of $a$, $b$ must be of the form $$a,b = 2^{\nu^2_{a,b}}7^{\nu^7_{a,b}}13^{\nu^{13}_{a,b}}23$$ with $0 \le \nu^p_{a,b}\le 1$ and $\nu^p_a + \nu^p_b = 1$ (as the gcd does not contain a factor $p$ and the lcm does), for $p \in \{2,7,13\}$. This leaves us with four possibilities (assuming $a \le b$):
• $a = 23$, $b = 2\cdot 7 \cdot 13 \cdot 23$,
• $a = 2 \cdot 23$, $b = 7 \cdot 13 \cdot 23$,
• $a = 7 \cdot 23$, $b = 2 \cdot 13 \cdot 23$,
• $a = 13 \cdot 23$, $b = 2 \cdot 7 \cdot 23$.