If the square of sum of three positive consecutive natural numbers exceeds the sum of their squares by 292, then what is the largest of the three numbers?
8
Let the three consecutive numbers be \(n, n+1, n+2\). Sum = \(3n+3\); sum of squares = \(3n^2 + 6n + 5\).
The condition \((3n+3)^2 - (3n^2+6n+5) = 292\) gives:
\[9n^2 + 18n + 9 - 3n^2 - 6n - 5 = 292\]
\[6n^2 + 12n - 288 = 0 \implies n^2 + 2n - 48 = 0\]
Factoring: \((n+8)(n-6)=0\). Since n is a positive natural number, \(n = 6\). The numbers are 6, 7, 8, so the largest is 8.
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