Answer
350.7k+ views
Hint: Here, first of all, let the two numbers be x and y. Now, according to the property, the product of two natural numbers will be equal to the product of its LCM and GCD. So, therefore, we will get an equation and on solving that equation, we will get our two numbers.
Complete step-by-step solution:
In this question, we are given that the least common multiple (LCM) of two numbers is 78 and the greatest common divisor (GCD) is 13.
Let these two numbers be x and y.
Now, we have a property that the product of two natural numbers x and y is equal to the product of its LCM and GCD. Therefore, we can say that for two natural numbers x and y,
$ \Rightarrow x \times y = GCD\left( {x,y} \right) \times LCM\left( {x,y} \right)$
Here, we have GCQ equal to 13 and the LCM as 78. Therefore, substituting these values in above equation, we get
$\Rightarrow x \times y = 13 \times 78 \\
\Rightarrow x \times y = 1014 $
Now, on prime factorization 1014, we get
$ \Rightarrow x \times y = 2 \times 3 \times 13 \times 13$
Now, 13 is common divisor of both numbers x and y, we can write
$ \Rightarrow x \times y = \left( {2 \times 13} \right) \times \left( {3 \times 13} \right)$
OR
$ \Rightarrow x \times y = \left( {1 \times 13} \right) \times \left( {2 \times 3 \times 13} \right)$
Hence, now x will be equal to $\left( {2 \times 13} \right)$ or $\left( {1 \times 13} \right)$ and y will be equal to $\left( {3 \times 13} \right)$ or $\left( {2 \times 3 \times 13} \right)$. Therefore,
$ \Rightarrow x = 2 \times 13 = 26$ OR $ \Rightarrow x = 1 \times 13 = 13$
$ \Rightarrow y = 3 \times 13 = 39$ OR $ \Rightarrow y = 2 \times 3 \times 13 = 78$
Hence, the two natural numbers whose LCM is 78 and GCD is 13 are 26 and 39 or 13 and 78.
Note: Here, we can cross verify our answer by finding the LCM and GCD of 26 and 39.
First of all let us find LCM of 26 and 39.
$
\Rightarrow 26 = 2 \times 13 \\
\Rightarrow 39 = 3 \times 13 $
Therefore,
$ \Rightarrow LCM = 2 \times 3 \times 13 = 78$
Now, let us find the GCD of 26 and 78.
$
\Rightarrow 26 = 2 \times 13 \\
\Rightarrow 39 \Rightarrow 3 \times 13 $
Hence,
$ \Rightarrow GCD = 13$
Hence, our answer is correct.
Complete step-by-step solution:
In this question, we are given that the least common multiple (LCM) of two numbers is 78 and the greatest common divisor (GCD) is 13.
Let these two numbers be x and y.
Now, we have a property that the product of two natural numbers x and y is equal to the product of its LCM and GCD. Therefore, we can say that for two natural numbers x and y,
$ \Rightarrow x \times y = GCD\left( {x,y} \right) \times LCM\left( {x,y} \right)$
Here, we have GCQ equal to 13 and the LCM as 78. Therefore, substituting these values in above equation, we get
$\Rightarrow x \times y = 13 \times 78 \\
\Rightarrow x \times y = 1014 $
Now, on prime factorization 1014, we get
$ \Rightarrow x \times y = 2 \times 3 \times 13 \times 13$
Now, 13 is common divisor of both numbers x and y, we can write
$ \Rightarrow x \times y = \left( {2 \times 13} \right) \times \left( {3 \times 13} \right)$
OR
$ \Rightarrow x \times y = \left( {1 \times 13} \right) \times \left( {2 \times 3 \times 13} \right)$
Hence, now x will be equal to $\left( {2 \times 13} \right)$ or $\left( {1 \times 13} \right)$ and y will be equal to $\left( {3 \times 13} \right)$ or $\left( {2 \times 3 \times 13} \right)$. Therefore,
$ \Rightarrow x = 2 \times 13 = 26$ OR $ \Rightarrow x = 1 \times 13 = 13$
$ \Rightarrow y = 3 \times 13 = 39$ OR $ \Rightarrow y = 2 \times 3 \times 13 = 78$
Hence, the two natural numbers whose LCM is 78 and GCD is 13 are 26 and 39 or 13 and 78.
Note: Here, we can cross verify our answer by finding the LCM and GCD of 26 and 39.
First of all let us find LCM of 26 and 39.
$
\Rightarrow 26 = 2 \times 13 \\
\Rightarrow 39 = 3 \times 13 $
Therefore,
$ \Rightarrow LCM = 2 \times 3 \times 13 = 78$
Now, let us find the GCD of 26 and 78.
$
\Rightarrow 26 = 2 \times 13 \\
\Rightarrow 39 \Rightarrow 3 \times 13 $
Hence,
$ \Rightarrow GCD = 13$
Hence, our answer is correct.
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