Codebymath.com - Online coding lessons using sum of consecutive squares

Lesson goal: Sum of Three Consecutive Squares

Previous: Divisibility of numbers $<1000$ | Home | Next: Father and Son Age Puzzle

Take any three consecutive integers: $n - 1$, $n$, and $n + 1$, where $n$ is the middle integer. If we square each integer and add them together: $$S = (n - 1)^2 + n^2 + (n + 1)^2$$ $$(n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n^2 + 2$$ In the 1970 MAA High School Mathematics Contest (Problem #4), students were asked to determine the divisibility of numbers in this set $S$: Can $3n^2 + 2$ ever be divisible by 2? By 3? By 5? By 7? By 11? Using a for-loop and an if-statement with the modulo operator (%), we can test these divisibility properties over a range of numbers!

sum = (n - 1)^2 + n^2 + (n + 1)^2
if sum % 3 == 0 then
print("Divisible by 3!")
end

Move the mouse over a dotted box for more information.

Notice the algebraic formula: $S = 3n^2 + 2 = 3(n^2) + 2$. When dividing by 3, the remainder is always 2, regardless of what integer $n$ is! This means that no member of $S$ is ever divisible by 3. But what about 11? Let's check with code.

Now you try. Replace `????` in both `if` conditions to test divisibility by 3 and 11. Run the code to verify that no sum is divisible by 3, but some (like $n=5$: $4^2 + 5^2 + 6^2 = 77$) are divisible by 11!

Type your code here:

See your results here:

This code will not run yet because the if conditions contain ????. Replace ???? with the proper modulo tests using sum % 3 == 0 and sum % 11 == 0.

Lesson goal: Sum of Three Consecutive Squares

Now you try. Replace ???? in both if conditions to test divisibility by 3 and 11. Run the code to verify that no sum is divisible by 3, but some (like $n=5$: $4^2 + 5^2 + 6^2 = 77$) are divisible by 11!

Share your code

Now you try. Replace `????` in both `if` conditions to test divisibility by 3 and 11. Run the code to verify that no sum is divisible by 3, but some (like $n=5$: $4^2 + 5^2 + 6^2 = 77$) are divisible by 11!