Find the LCM of the four numbers (3, 5, 8, 12)
3: 3
5: 5
8: 2 * 2 * 2
12: 2 * 2 * 3
LCM = 2 * 2 * 2 * 3 * 5
= 120
Answer: 120 years
Let x = smallest number
y = middle number
z = largest number
Equations:
sum of 3 numbers: x + y + z = 15
two times smallest is 1 less than largest: 2x = z - 1
sum of largest and smallest is 10: z + x = 10
z + x = 10 can be solved for z ⇒ z = 10 - x
substitute into 2x = z - 1 ⇒ 2x = 10 - x - 1
3x = 9
x = 3 (divided both sides of equation by 3)
Since z + x = 10 ⇒ z + 3 = 10 ⇒ z = 7
Substitute values for x and z into x + y + z = 15 ⇒ 3 + y + 7 = 15
y + 10 = 15
y = 5
answer: 3, 5, 7
Answer:
13 teams and 3 people left over
Step-by-step explanation:
55/4 would be 13 with a remainder of 3. That 3 would be the number of people left over. And 13 would be the number of groups of four available.
