2021 CMC 12A Problems/Problem 7

The following problem is from both the 2021 CMC 12A #7 and 2021 CMC 10A #9, so both problems redirect to this page.

Problem

It is known that every positive integer can be represented as the sum of at most $4$ squares. What is the sum of the $2$ smallest integers which cannot be represented as the sum of fewer than $4$ squares?

$\textbf{(A) } 22\qquad\textbf{(B) } 23\qquad\textbf{(C) } 24\qquad\textbf{(D) } 25\qquad\textbf{(E) } 26\qquad$

Solution

$\begin{align*} 1 &= 1^2 \\ 2 &= 1^2 + 1^2 \\ 3 &= 1^2 + 1^2 + 1^2 \\ 4 &= 2^2 \\ 5 &= 1^2 + 2^2 \\ 6 &= 1^2 + 1^2 + 2^2 \\ 7 \\ 8 &= 2^2 + 2^2 \\ 9 &= 3^2 \\ 10 &= 1^2 + 3^2 \\ 11 &= 1^2 + 1^2 + 3^2 \\ 12 &= 2^2 + 2^2 + 2^2 \\ 13 &= 2^2 + 3^2 \\ 14 &= 1^2 + 2^2 + 3^2 \\ 15 \\ \end{align*}$

We cannot find a solution in fewer than $4$ squares for $7$ and $15$ , so the answer is $7+15=\boxed{\textbf{(A) } 22}$ .

2021 CMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 12 Problems and Solutions

2021 CMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2021 CMC 12A Problems/Problem 7

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