Browse · harp Print → jmc prealgebra junior Problem 4(299)+3(299)+2(299)+298= (A) 2889 (B) 2989 (C) 2991 (D) 2999 Solution — click to reveal We can make use of the distributive property as follows: 4(299)+3(299)+2(299)+298=4(299)+3(299)+2(299)+1(299)−1=(4+3+2+1)(299)−1=10(299)−1=2989 2989 Final answer B ← Previous problem Next problem →