Question 53220
Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91. 
LET N,N+1,N+2,N+3 BE THE 4 INTEGERS
SUM OF 2 GREATER INTEGERS = N+2+N+3=2N+5
TWICE THE SUM = 2(2N+5)=4N+10
3 TIMES THE FIRST = 3N
4N+10=3N+91
N=81
SO THE 4 INTEGERS ARE 81,82,83,84