1. **State the problem:** We need to find the smallest positive integer $m$ such that both 28 and 90 are factors of $m$. 2. **Formula and rules:** If two numbers are factors of $m$, then $m$ must be a multiple of their least common multiple (LCM). 3. **Find the prime factorization:** - $28 = 2^2 \times 7$ - $90 = 2 \times 3^2 \times 5$ 4. **Calculate the LCM:** The LCM is found by taking the highest powers of all prime factors present: $$\text{LCM}(28, 90) = 2^2 \times 3^2 \times 5 \times 7$$ 5. **Calculate the value:** $$2^2 = 4$$ $$3^2 = 9$$ $$4 \times 9 = 36$$ $$36 \times 5 = 180$$ $$180 \times 7 = 1260$$ 6. **Conclusion:** The smallest possible value of $m$ is $1260$ because it is the smallest number divisible by both 28 and 90.