Question

Find 3 consecutive positive integers whose product is equal to 16 times their sum

Answer 1

let m,(m + 1),(m + 2) be the consecutive integers.

As per given condition,
m(m + 1)(m+2) = 16(m+m+1+m+2)
(m²+2m)(m+1) = 16(3m+3)
(m²+2m)(m+1) = 48(m+1)
m² + 2m = 48
m² + 2m - 48 = 0
m² + 8m - 6m - 48 = 0
m(m + 8) - 6(m + 8) = 0
(m + 8)(m - 6) = 0

m = -8 OR. m = 6

(m) is positive integer, so m≠-8, thus m=6

THEREFORE, THE POSITIVE INTEGERS ARE
m = 6
m + 1 = 6 + 1 = 7
m + 2 = 6 + 2 = 8

i.e., 6,7,8 are the consecutive positive integers.

Answer 2

Answer:

Step-by-step explanation:

.As per given condition,

m(m + 1)(m+2) = 16(m+m+1+m+2)

(m²+2m)(m+1) = 16(3m+3)

(m²+2m)(m+1) = 48(m+1)

m² + 2m = 48

m² + 2m - 48 = 0

m² + 8m - 6m - 48 = 0

m(m + 8) - 6(m + 8) = 0

(m + 8)(m - 6) = 0

m = -8 OR. m = 6

(m) is positive integer, so m≠-8, thus m=6

THEREFORE, THE POSITIVE INTEGERS ARE

m = 6

m + 1 = 6 + 1 = 7

m + 2 = 6 + 2 = 8