⇔ We know that:
↔ Consider the smallest integer = a
↔ So, the next 2 integers will be [tex](a+2)[/tex] and [tex](a+4)[/tex]
⇔ Solution:
⇒ [tex]a+(a+2)+(a+4)=216[/tex]
⇒ [tex]3a=216-6[/tex]
⇒ [tex]3a=210[/tex]
⇒ [tex]a= \frac{210}{3}=\boxed{70} [/tex]
⇔ So, the largest possible value of one of these integers is:
∴ [tex]a+4=70+4=\boxed{\boxed{74}}[/tex]
