To find the total number of integers given that their GCF (Greatest Common Factor) is 3 and LCM (Least Common Multiple) is 180, with one of the numbers being 45, you can use the relationship between GCF, LCM, and the product of the numbers.
Using the formula:
\[ \text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b \]
Let the two numbers be \( a \) and \( b \). Given:
- \( \text{GCF}(a, b) = 3 \)
- \( \text{LCM}(a, b) = 180 \)
- Assume \( a = 45 \).
We need to find \( b \).
Plugging in the values, we have:
\[ 3 \times 180 = 45 \times b \]
Calculating the left-hand side:
\[ 540 = 45 \times b \]
Solving for \( b \):
\[ b = \frac{540}{45} = 12 \]
Thus, the two numbers are \( 45 \) and \( 12 \).
Now let's verify that their GCF is 3 and their LCM is 180:
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The GCF of \( 45 \) and \( 12 \):
- The prime factorization of \( 45 \) is \( 3^2 \times 5 \).
- The prime factorization of \( 12 \) is \( 2^2 \times 3 \).
- The common factor is \( 3^1 \), so \( \text{GCF} = 3 \).
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The LCM of \( 45 \) and \( 12 \):
- The LCM is found by taking the highest power of each prime factor:
- From \( 45 \): \( 3^2 \) and \( 5^1 \)
- From \( 12 \): \( 2^2 \) and \( 3^1 \)
- Thus, \( \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180 \).
- The LCM is found by taking the highest power of each prime factor:
Since both the GCF and LCM conditions are satisfied, the total number of integers in this case is two: \( 45 \) and \( 12 \).