Question

Enzo and Beatriz are playing games at their local arcade. Incredibly, Enzo wins 55 tickets from every game, and Beatriz wins 1111 tickets from every game. When they stopped playing games, Enzo and Beatriz had won the same number of total tickets. What is the minimum number of games that Enzo could have played?

Answer 1

Answer:

101

Step-by-step explanation:

Note that

• $55=5\cdot 11;$
• $1111=11\cdot 101.$

Here these two numbers have the common divisor 11. Find the least common multuply:

$LCM(55,1111)=11\cdot 5\cdot 101=5555.$

This means that  the minimum number of games that Enzo could have played will be

$\dfrac{5555}{55}=101.$