Question

Three consecutive odd integers have the property that the sum of 4 more than
the smallest and one-third of the largest is 71 more than the second number.
Find the integers.

Answer 1

let the number be
a,a+2,a+4
then it is given that 4+a+(a+4)/3=71+a+2
so, by solving for a, we get
a=203
so numbers are
203,205,207