Question

If n and m are two different prime numbers that are both less than 10 , what is the greatest possible value of the greatest common factor of 4 n m and 5 m n ?

Answer 1

To find the greatest possible value of the greatest common factor (GCF) of the expressions \(4nm\) and \(5mn\), where \(n\) and \(m\) are different prime numbers less than 10, we start by expressing the GCF mathematically.

\[
\text{GCF}(4nm, 5mn) = mn \cdot \text{GCF}(4, 5)
\]

Next, we compute \(\text{GCF}(4, 5)\):
- The prime factorization of \(4\) is \(2^2\).
- The prime factorization of \(5\) is \(5^1\).

Since \(4\) and \(5\) have no common factors, we find:

\[
\text{GCF}(4, 5) = 1
\]

This simplifies our GCF calculation:

\[
\text{GCF}(4nm, 5mn) = mn \cdot 1 = mn
\]

To maximize the value of \(mn\), we need to select two different primes \(n\) and \(m\) that are both less than 10. The prime numbers less than 10 are \(2, 3, 5, 7\).

Now, we calculate \(mn\) for all combinations of these primes:

- For \(n = 2\) and \(m = 3\):
\[
mn = 2 \cdot 3 = 6
\]

- For \(n = 2\) and \(m = 5\):
\[
mn = 2 \cdot 5 = 10
\]

- For \(n = 2\) and \(m = 7\):
\[
mn = 2 \cdot 7 = 14
\]

- For \(n = 3\) and \(m = 5\):
\[
mn = 3 \cdot 5 = 15
\]

- For \(n = 3\) and \(m = 7\):
\[
mn = 3 \cdot 7 = 21
\]

- For \(n = 5\) and \(m = 7\):
\[
mn = 5 \cdot 7 = 35
\]

Now, we find the maximum value from all the products:

- \(6\) (from \(2 \cdot 3\))
- \(10\) (from \(2 \cdot 5\))
- \(14\) (from \(2 \cdot 7\))
- \(15\) (from \(3 \cdot 5\))
- \(21\) (from \(3 \cdot 7\))
- \(35\) (from \(5 \cdot 7\))

The greatest value of \(mn\) is \(35\), which occurs when \(n = 5\) and \(m = 7\).

Thus, the greatest possible value of the GCF of \(4nm\) and \(5mn\) is:

\[
\boxed{35}
\]