Question

Find three consecutive odd integers such that 5 times the sum of all three is 72 more than the product of the first and second integers

Answer 1

The three consecutive odd integers are 7, 9 and 11

Solution:

Let the three consecutive odd integers be n, n + 2, n + 4

Given that,

5 times the sum of all three is 72 more than the product of the first and second integers

Which means,

5 times sum of n, n + 2, n + 4 = 72 + product of first and second integer

Writing it mathematically we get,

$5(n + n + 2 + n + 4) = 72 + n(n+2)\\\\Solve\ the\ equation\ for\ x\\\\5(3n + 6) = 72 + n^2 + 2n\\\\15n + 30 = 72 + n^2 + 2n\\\\n^2 + 2n -15n + 72-30 = 0\\\\n^2 -13n + 42 = 0$

Factor the above quadratic equation

Split the middle term

$n^2-6n - 7n + 42 = 0\\\\Group\ the\ terms\\\\(n^2-6n)-(7n - 42) = 0\\\\Factor\ out\ n\ from\ first\ group\ and\ factor\ out\ 7\ from\ second\ group\\\\n(n - 6) -7(n -6) = 0\\\\Factor\ out\ (n-6)\\\\(n-6)(n-7) = 0\\\\Equate\ to\ zero\\\\n = 6 \text{ or } n = 7$

Given that "n" is a odd integer

Therefore, choose n = 7

The three consecutive integers are:

n = 7

n + 2 = 7 + 2 = 9

n + 4 = 7 + 4 = 11

Thus the three consecutive odd integers are 7, 9 and 11