Find the least positive integer $m$ such that $lcm(15,m) = lcm(42,m)$. Here $lcm(a, b)$ is the least common multiple of $a$ and $b$.
To find the least positive integer \( m \) such that \( \text{lcm}(15, m) = \text{lcm}(42, m) \), we start by using the definition of the least common multiple (LCM) and the greatest common divisor (GCD). The LCM of two numbers \( a \) and \( b \) can be expressed as:

\[
\text{lcm}(a, b) = \frac{a \cdot b}{\text{gcd}(a, b)}
\]

Given the problem, we need:

\[
\text{lcm}(15, m) = \text{lcm}(42, m)
\]

Using the LCM formula, we have:

\[
\frac{15 \cdot m}{\text{gcd}(15, m)} = \frac{42 \cdot m}{\text{gcd}(42, m)}
\]

By canceling \( m \) from both sides (assuming \( m \neq 0 \)), we get:

\[
\frac{15}{\text{gcd}(15, m)} = \frac{42}{\text{gcd}(42, m)}
\]

Let \( \text{gcd}(15, m) = d_1 \) and \( \text{gcd}(42, m) = d_2 \). Then the equation becomes:

\[
\frac{15}{d_1} = \frac{42}{d_2}
\]

This implies:

\[
\frac{d_2}{d_1} = \frac{42}{15} = \frac{14}{5}
\]

Since \( d_1 \) and \( d_2 \) must be divisors of 15 and 42 respectively, and their ratio must be \( \frac{14}{5} \), we need to find \( d_1 \) and \( d_2 \) such that:

\[
d_2 = \frac{14}{5} d_1
\]

The only possible values for \( d_1 \) and \( d_2 \) that satisfy this ratio and are divisors of 15 and 42 are \( d_1 = 5 \) and \( d_2 = 14 \). Therefore:

\[
\text{gcd}(15, m) = 5 \quad \text{and} \quad \text{gcd}(42, m) = 14
\]

Next, we need to find the smallest \( m \) such that \( \text{gcd}(15, m) = 5 \) and \( \text{gcd}(42, m) = 14 \). 

Since \( \text{gcd}(15, m) = 5 \), \( m \) must be of the form \( 5k \) where \( k \) is an integer not divisible by 3 (since 15 = 3 * 5).

Since \( \text{gcd}(42, m) = 14 \), \( m \) must be of the form \( 14j \) where \( j \) is an integer not divisible by 3 (since 42 = 2 * 3 * 7).

Combining these, \( m \) must be a multiple of both 5 and 14, but not divisible by 3. The least common multiple of 5 and 14 is:

\[
\text{lcm}(5, 14) = 70
\]

Thus, the smallest \( m \) that satisfies both conditions is 70.

The final answer is \( \boxed{70} \).