Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.8 AMC 10/12 Find the answer South Africa

The sum of the squares of 3 consecutive positive integers is 770. The largest of these integers is

Pick one

Solution

If we call the three numbers $n-1$ , $n$ , $n+1$ , then $(n-1)^2 + n^2 + (n+1)^2 = 770$ , which simplifies to $3n^2 + 2 = 770$ . Therefore $n^2 = 768/3 = 256 = 16^2$ , so $n = 16$ (since $n > 0$ ), and the largest number is $n+1 = 17$ .

Without doing any algebra, it is clear that $n^2 \approx 770/3 = 256\frac{2}{3}$ , and the nearest perfect square is 256.

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