Asked by sarah
what is the smallest of three consecutive numbers when squared add up to 365
Answers
Answered by
oobleck
If the smallest is x, then
x^2 + (x+1)^2 + (x+2)^2 = 365
x=10
Or, since the three numbers are consecutive, they are all almost equal. So, if the middle number is x, then
3x^2 ≈ 365
x^2 ≈ 121
x = 11
That makes the smallest equal to 10.
x^2 + (x+1)^2 + (x+2)^2 = 365
x=10
Or, since the three numbers are consecutive, they are all almost equal. So, if the middle number is x, then
3x^2 ≈ 365
x^2 ≈ 121
x = 11
That makes the smallest equal to 10.
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