Minimum Quotient

The value of $m(x)$ can be found by multiplying all the prime numbers less than or equal to $x$ , and raising each prime number $p$ to the highest exponent $n$ such that $p^n \leq x$ .

Therefore,

$m(120) = 2^6 \cdot 3^4 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \dots 113$

$m(130) = 2^7 \cdot 3^4 \cdot 5^3 \cdot 7^2 \cdot 11^2 \cdot 13 \dots 113 \cdot 127$

So $\frac{m(130}{m(120)} = \frac{2^7 \cdot 3^4 \cdot 5^3 \cdot 7^2 \cdot 11^2 \cdot 13 \dots 113 \cdot 127}{2^6 \cdot 3^4 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \dots 113} = 2 \cdot 5 \cdot 11 \cdot 127 = \boxed{13970}$ .

$Mathematica$

LCM@@Range@130/LCM@@Range@120

returns $13970$

The smallest number that evenly divides $m(x)$ is the $\text{lcm}$ from $1$ to $x$ and $\text{lcm(1.2.3,...,x)}$ is the product of all primes numbers less than or equal $x$ ie $\text{lcm(1 to x )}$ is $\small (p_1^{q_1}\leq x)\cdot\,(p_2^{q_2}\leq x)\cdot(p_3^{q_3}\leq x)\cdots (p_n^{q_n}\leq x )$

$\therefore \dfrac{\text{lcm(1,2,3,....,130)}}{\text{lcm(1,2,3,4,.....120)}} = \dfrac{2^7\cdot 3^4\cdot 5^3\cdot 7^2\cdot 11^2}{2^6\cdot 3^4 \cdot 5^2 \cdot 7^2 \cdot 11}\left(\dfrac{13\cdot 17\cdots 23\cdots 113}{13\cdot 17\cdots23 \cdots 113}\right) \cdot 127 \\ \implies \dfrac{m(130)}{m(120)} = 2\cdot 5\cdot 11\ 127 = 13970$

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Note: In case of $\text{lcm}(1,2,3,...130)$ , $128(2^7) , 125(5^3),81(3^4),49(7^2), 121(11^2)$ appears multiple times and primes between 11 and 130 appears only a time for composite numbers between them. On the same way lcm from 1 to 120 can also be determined.