How can the whole number 125 be expressed as a product of prime numbers with exponents?

Answer 1

To express the whole number 125 as a product of prime numbers with exponents, we first need to factor 125 into its prime components.

We can start by dividing 125 by the smallest prime number, which is 2. However, since 125 is odd, it is not divisible by 2. Next, we try dividing by 3, which also does not work (125 is not divisible by 3).

Next, we check 5:

• \(125 \div 5 = 25\)

Now we factor 25, which can be divided by 5 again:

• \(25 \div 5 = 5\)

Finally, we end up with:

• \(5 \div 5 = 1\)

So, we see that 125 can be factored as follows:

\[ 125 = 5 \times 5 \times 5 = 5^3 \]

Therefore, the expression of 125 as a product of prime numbers with exponents is:

\[ 125 = 5^3 \]

Answer 2

How can the whole number 378 be expressed as a product of prime numbers with exponents?

Answer 3

To express the whole number 378 as a product of prime numbers with exponents, we will perform prime factorization.

1. Start by dividing 378 by the smallest prime number, which is 2: \[ 378 \div 2 = 189 \] So, \(378 = 2 \times 189\).

2. Next, we factor 189. Since 189 is odd, we cannot divide it by 2. Let's try dividing by 3: \[ 189 \div 3 = 63 \] Thus, \(189 = 3 \times 63\).

3. Now, we factor 63. We can again divide by 3: \[ 63 \div 3 = 21 \] So, \(63 = 3 \times 21\).

4. We continue with 21 and divide by 3 once more: \[ 21 \div 3 = 7 \] Thus, \(21 = 3 \times 7\).

5. Finally, we have \(7\) which is a prime number.

Putting it all together, we have:

\[ 378 = 2 \times 3 \times 3 \times 3 \times 7 \]

This can be rewritten using exponents for repeated prime factors:

\[ 378 = 2^1 \times 3^3 \times 7^1 \]

Thus, the prime factorization of 378 is:

\[ 378 = 2^1 \times 3^3 \times 7^1 \]