# The school cafeteria serves tacos every sixth day and cheeseburgers every eight day. If tacos and cheeseburgers are both on today's menu, how many days will it be before they are both on the menu again?

Nov 11, 2017

24 days

#### Explanation:

If we consider today as Day 0, then
Days with tacos: 6, 12, 18, 24, ...
Days with cheeseburgers: 8, 16, 24, ...
It can be seen that after 24 days, both will be on the menu again.

In fact, this utilizes LCM (lowest common multiple) in calculations. By prime factorization,
$6 = 2 \cdot 3$
$8 = 2 \cdot 2 \cdot 2$

As both of them have a 2, we can take the two out and count it once. Therefore,

$L C M \left(6 , 8\right) = 2 \cdot 3 \cdot 2 \cdot 2 = 24$,
Where the first 2 is the common factor, 3 comes from the factor of 6 and the 2*2 from 8.
In this way, we can find the number of days, which is 24.

Nov 11, 2017

Every 24th day.

#### Explanation:

Find the L.C.M. OF 6 & 8. It will be 24.

Hence both the menus will be together every 24th day.

Nov 11, 2017

Counting numbers as objects. The object of 8 has within it the object of 6 and part of another 6.

24

#### Explanation:

Although there will be a greater count of 6's for a given count of 8 only particular ones of the 6's will coincide with particular ones of the 8's.

Sounds a bit obvious but for every 8 we have a 6 plus part of another 6. In that we have $6 + 2 = 8$

So if we accumulate these we have.

$\textcolor{w h i t e}{\text{1}} 6 + 2 = 8$
$\textcolor{w h i t e}{\text{1}} 6 + 2 = 8$
$\textcolor{w h i t e}{\text{1")ul(6+2=8 larr" Add}}$
$18 + 6 = 24$
$\textcolor{w h i t e}{\text{1111}} \textcolor{red}{\uparrow}$
$\textcolor{red}{\text{Coincides when all the 'bits' of a 6 add up to give another 6}}$

We have a count of 4 at 6 and a count of 3 at 8.