Question

The sum of four consecutive integers is 1290 the greatest of them is

Answer 1

324

Step-by-step explanation:

let x,x+1,x+2,x+3 be the four consecutive integers

(x)+(x+1)+(x+2)+(x+3)=1290

4x+6=1290

4x=1290-6

4x=1284

x=321

therefore

x+1= 321+1= 322

x+2 = 321+2= 323

x+3 = 321+3= 324