Algebra-9 (CAT-2008)

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals to the total of the three original integers. Which of the following best describes the minimum, say m of these three numbers?

(1)1 ≤ m ≤ 3 (2)4 ≤ m ≤ 6 (3)7 ≤ m ≤ 9 (4)10 ≤ m ≤ 12 (5) 13 ≤ m ≤ 15
Solution follows here:

Solution:

Let the numbers be n-1,n,n+1

(n-1)+n²+(n+1)³ = (n-1+n+n+1)² => n-1+n²+n³+1+3n+3n² = (3n)²

n³+4n²+4n = 9n² => n³-5n²+4n = 0 => n(n²-5n+4) = 0

n(n-1)(n-4) = 0 => n= 0 or 1 or 4

for n=0, the numbers are: -1,0,1  this is out as all the numbers should be positive

for n=1, the numbers are: 0,1,2  this is out as all the numbers should be positive (‘0’ can’t be taken as positive)

for n=4, the numbers are: 3,4,5  this is correct and the minimum number out of the three is ‘3’

For ‘3’, the best option to select is (1)

Answer (1)