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South African Mathematics Olympiad

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−1n-1, nn, n+1n+1, then (n−1)2+n2+(n+1)2=770(n-1)^2 + n^2 + (n+1)^2 = 770, which simplifies to 3n2+2=7703n^2 + 2 = 770. Therefore n2=768/3=256=162n^2 = 768/3 = 256 = 16^2, so n=16n = 16 (since n>0n > 0), and the largest number is n+1=17n+1 = 17.

Without doing any algebra, it is clear that n2≈770/3=25623n^2 \approx 770/3 = 256\frac{2}{3}, and the nearest perfect square is 256.

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