Question

If the sum of three consecutive odd integers is at MOST 216, what is the largest possible value of one of these integers?

a) 69

b) 71

c) 74

d) 73

Answer 1

⇔ We know that:
↔ Consider the smallest integer = a
↔ So, the next 2 integers will be $(a+2)$ and $(a+4)$

⇔ Solution:
⇒ $a+(a+2)+(a+4)=216$
⇒ $3a=216-6$
⇒ $3a=210$
⇒ $a= \frac{210}{3}=\boxed{70}$

⇔ So, the largest possible value of one of these integers is:
∴ $a+4=70+4=\boxed{\boxed{74}}$