Given `4725=3^a5^b7^c ,` find the integral values of `a ,\ b\ a n d\ c` the value of `2^(-a)3^b7^c`

Explanation

To solve the problem given \( 4725 = 3^a \cdot 5^b \cdot 7^c \), we need to factor 4725 into its prime factors and then find the values of \( a \), \( b \), and \( c \).

### Step 1: Factorize 4725
To factor 4725, we start by dividing it by the smallest prime numbers.

1. **Dividing by 5**: Since 4725 ends with a 5, it is divisible by 5.
\[
4725 \div 5 = 945
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