[email protected] wrote:i cant understand the explanation...can someone please elaborate
Hi Roopali, it is simple.. Let me give another try..
Given that
(3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x )
This can be rewritten as
3^4 * 5^6 * 7^3 = 7^n * 5^n * x
Any number can be written as a product of primes.
When two numbers are same, the primes involved in their products and their exponents should be same.
Example: 20 = 4 * 5 = 2^2 * 5
20 can also be written as 10*2 = 2*5*2 = 2^2 * 5
Hence, X in the above problem should have a multiple of 3^4. Which means x = (3^4*k) for some k
Now, Let us rewrite..
3^4 * 5^6 * 7^3 = 7^n * 5^n * (3^4*k)
There are 6 fives and 3 sevens on LHS
Now, what is the maximum value n can take??
Suppose n takes 6, then on the right hand side we will get 6 sevens which is wrong.
We can have a maximum of 3 sevens on RHS.
Hence n can take only 3 at maximum.
Eqtn will then look like
3^4 * 5^6 * 7^3 = 7^3 * 5^3 * (3^4*5^3)
Let me know if you have any troubles understanding this.
