1 2 and 3
7+8+9=24 7x8x9=504
lol I need to know this too.
The first three positive integers, 1, 2, and 3, satisfy this condition.
Three consecutive integers have a sum of 12. What is the greatest of these integers?
Not sure what thress is. If three, then there is no answer since the sum (or product) of any three consecutive integers must be divisible by 3.
-- Their sum and difference both have the same sign that the two integers have. -- Their product and quotient are both positive.
Their product is 143.
The product of the two integers is -80.
The answer would be 10 12 and 14... 14 x 3 = 42 and 2(10 + 12) = 44. So the product of the largest integer and three is two less than twice the sum of the lower integers.
The integers are 5 and 7.
There is no set of three consecutive integers whose sum is 71.
No integers work