How many positive integers a less than 100 have a corresponding integer b divisible by 3 such that the roots of x^2-ax+b=0 are consecutive positive integers?

There are 24 such numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95

Step-by-step explanation:

Solving, we have

(x-y)(x-(y-1))=x^2-(2×y-1)x+y^2-y=x^2-ax+b comparison shows that

(2y-1)=a and y^2-y =b or

y(y-1)=b

This gives the following qualifying numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95

answer: 414,377,008

step-by-step explanation:

just multiply it