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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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    Okay, I'd edit the question.2012-12-12

1 Answers 1

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$\def\lcm{\operatorname{lcm}}$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 $a \le b$):

  • $a = 23$, $b = 2\cdot 7 \cdot 13 \cdot 23$,
  • $a = 2 \cdot 23$, $b = 7 \cdot 13 \cdot 23$,
  • $a = 7 \cdot 23$, $b = 2 \cdot 13 \cdot 23$,
  • $a = 13 \cdot 23$, $b = 2 \cdot 7 \cdot 23$.

So, as far as I see, the bigger number cannot be determined exactly.