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- Wait — correction: in earlier computation, we had $4 \cdot 3 \cdot 5 \cdot 7 \cdot 17 \cdot 119$, but $119 = 7 \cdot 17$, and $119$ has repeated factor 7, so $7^1$ is sufficient.
- But $1,2,3,5,7,13,109$: all distinct primes → LCM = product of all distinct prime factors = $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 109$
- Indeed:
- Can we do better? Try including $17$? Suppose $1, 2, 3, 5, 17, 19, 92$: sum = $1+2+3+5+17+19+92 = 140$
- LCM includes $2^2$, $3$, $5$, $17$, $19$, $92 = 4 \cdot 23$, so $23$ introduced.
- LCM = $ \text{LCM}(4,3,5,17,19,23) = 4 \cdot 3 \cdot 5 \cdot 17 \cdot 19 \cdot 23 $
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