The exponent of 3 in the prime factorisation of 243 is :
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Step-by-Step Solution
Step 1: Understand Prime Factorization
Prime factorization is the process of breaking down a composite number into its prime factors. A prime factor is a prime number that divides the given number exactly. For example, the prime factors of 12 are 2, 2, and 3, so 12=22×3.
Step 2: Start dividing by the smallest prime number
To find the prime factors of 243, we start by dividing it by the smallest prime number, which is 3. Since the sum of the digits of 243 (2+4+3=9) is divisible by 3, 243 is divisible by 3.
Step 3: Continue dividing the quotient by 3
We continue the process by dividing the quotient, 81, by 3 again. The sum of the digits of 81 (8+1=9) is divisible by 3, so 81 is divisible by 3.
Step 4: Repeat the division
We repeat the division with the new quotient, 27. Since 27 is divisible by 3, we divide it by 3.
Step 5: Final division
Finally, we divide 9 by 3, which gives us 3. Since 3 is a prime number, we stop here.
Step 6: Write the prime factorization
By collecting all the prime factors, we can write 243 as a product of powers of its prime factors. We found that 3 was used as a factor 5 times.
Step 7: Identify the exponent of 3
From the prime factorization 243=35, the exponent of 3 is 5.