Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 1.9 Junior Find the answer

If $\sqrt{25-\sqrt{n}}=3$ , what is the value of $n$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since $\sqrt{25-\sqrt{n}}=3$ , then $25-\sqrt{n}=9$ . Thus, $\sqrt{n}=16$ and so $n=16^{2}=256$ .

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