# Dining tables

In the dining room are tables with 4 chairs, 6 chairs, 8 chairs.
How many diners must be at least to be occupy all tables (chairs) and diners are more than 50?

Result

n =  72

#### Solution:

$LCM(4, 6, 8) = 2^3 \cdot 3 = 24 \ \\ n > 50 \ \\ 24 k > 50 ; k \in Z \ \\ k > 2.08333 \ \\ k = 3 \ \\ \ \\ n = 24 k = 24\cdot 3 = 72$

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