Three cousins have ages that are consecutive integers. The product of the two older cousins

Question

Annie Levasseur · Accepted Answer

Step 1  x,x+1,x+2  Step 2  Product of 2 oldes cousins ages are twelve man 6 times the sum of younges two cousine age, so  (x+2)×(x+1)=6(x+x+1)−12  x2+3x+2=6(2x+1)−12  x2+3x+2=12x+6−12  x2+3x+2−12x=−6  x2−9x+8=0  x2−8x−x+8=0  x(x−8)−1(x−8)=0  (x−1)(x−8)=0  x−1=0 x−8=0  x=1 x=8  1,2,3 8,9,10  are the ages of 3 cousins

Heather Fulton · Accepted Answer

n,n+1,n+2   (n+1)(n+2)=6(n+n+1)−12   n2+3n+2=12n+6−12   n2+3n+2=12n−6   n2−9n+8=0   n2−n−8n+8=0   n(n−1)−8(n−1)=0   (n−8)(n−1)=0   So there are actually two solutions that satisfy the conditions for the cousins

nick1337 · Accepted Answer

Step 1
Given that three cousins have ages that are consecutive integers.
Let consecutive intergers be n-1, n, n+1
I.e. smallest cousine have age (n-1)

Middle age cousine have age n and
Eldest cousine have age (n+1)
Step 2
It is given that the product of the two older cousin's ages is twelve less than six times the sum of the younger two cousin's ages.
Mathematically,
$n (n + 1) = 6 [n + (n - 1)] - 12$
$n^{2} + n = 6 (2 n - 1) - 12$
$n^{2} + n = 12 n - 6 - 12$
$n^{2} - 11 n + 18 = 0$
$n^{2} - 9 n - 2 n + 18 = 0$
$n (n - 9) - 2 (n - 9) = 0$
$(n - 2) (n - 9) = 0$
Therefore, (n - 2) = 0 or (n - 9) = 0
It gives, n = 2 or n = 9
Step 3
Therefore for two different values of n, there are actually two solutions that satisfy the conditions for the cousin's ages...
Case 1 :
n = 2 gives ages of cousins as 1, 2 and 3 years.
Case 2:
n = 9 gives ages of cousins as 8, 9 and 10 years.

Three cousins have ages that are consecutive integers. The product

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