3T solutions #
Least common multiples #
Consider \( a = 2^2 \cdot 5^4 \cdot 7 \cdot 41^{10} \) and \( b = 5^2 \cdot 19 \cdot 41 \cdot 47^2 \) .
If \( M \) is a common multiple of \( a \) and \( b \) , that means that \( a | M \) and also \( b | M \) .
So what can we say about the pf of \( M \) ?
The closed form of \( M \) :
\[\begin{aligned} M &= 2^i \cdot 5^j \cdot 7^r \cdot 19^k \cdot 41^s \cdot 47^t \cdot u \end{aligned}\]where
\[\begin{aligned} i &\geq 2 \\ j &\geq 4 \\ k &\geq 1 \\ r &\geq 1 \\ s &\geq 10 \\ t &\geq 2 \\ u &\in \mathbb{Z} \end{aligned}\]So the shortcut is the union of primes with the higher power in each number.
Example
\( a = 2 \cdot 5^3 \cdot 19^4 \)
and
\( b = 2^4 \cdot 19 \cdot 41^7 \)
So our \( \text{LCM} (a, b) = 2^4 \cdot 5^3 \cdot 19^4 \cdot 41^7 \)
Example
\( a = 2 \cdot 5^3 \)
and
\( b = 2^2 \cdot 5^4 \cdot 11^9 \)
So our \( \text{LCM} (a,b) = 2^2 \cdot 5^4 \cdot 11^9 \)
Notice that the least common multiple is \( b \) ( \( b \) is a multiple of \( a \) ).
Example
\( a = 2^2 \cdot 3^2 \cdot 7 \)
.
Find \( b \) such that \( \text{LCM} (a, b) = 2^2 \cdot 3^{15} \cdot 7 \cdot 19^3 \)
\( b = 2^2 \cdot 3^{15} \cdot 7 \cdot 19^3 \)