(i) 120,90 (ii) 50,60 (iii) 37,49
Let us apply fundamental therom of arthematic.
Example 3.
Consider the numbers of the form 4n where number is a natural number.Check wheather there is any value of n for which 4n ends with zero?
Solution:
If 4n is to end with zero for a nutural number n it should be divided by 2 and 5. This means that the prime fractorization 4n should contain the prime number 5 and 2.But it is not possible because 4n=(2)2n so 2 is the only prime factorization of 4n.Since 5 is not present in the prime factorization, there is no natural number n for which 4n ends with the digit zero.
You have already learnt how to find the HCF (Highest Common Factor) and LCM (Lowest Common Multiple) of teo positive integers using the Fundamental Therom of Arthematic in earlier classes, without relizing it!This method is also called the prime factorization method. Let us recall this method through the following example.
Example 4.
Find the HCF and LCM of 12 and 18 by the prime factorization method.
Solution:
we have
Note that HCF (12,18)=21×31=Product of the smallest power of each common prime factor of the numbers.
LCM (12,18)=22×32=Product of the greatest powder of each prime factor of the numbers.
From the example above to might you may have notice that HCF (12,18)×LCM [12,18]=12×18.In fact,we can verify that for any two positive integers a and b, HCF (a,b) × LCM [a,b]=a×b.We can use this result to find the LCM of two positive integers, if we have already found the HCF of two positive integers.
| Find the HCF and LCM of the following given pair of numbers by the prime factorization method. (i) 120,90 (ii) 50,60 (iii) 37,49 |
| Show that 3n × 4m cannot end with the digit 0 or 5 for any natural numbers n and m |