How many factors of 729 are perfect squares?
4
4.
Step 1 — Prime factorisation: \(729 = 3^{6}\), so every factor has the form \(3^{k}\) with \(k = 0, 1, \ldots, 6\) (seven factors in all).
Step 2 — A factor is a perfect square exactly when its exponent is even: \(k = 0, 2, 4, 6\).
Step 3 — These give \(3^{0} = 1,\ 3^{2} = 9,\ 3^{4} = 81,\ 3^{6} = 729\), i.e. four perfect-square factors.
Hence, 4.
Note: The official RRB answer key marks 3 for this question. Counting every factor of 729, including 1 and 729 itself, there are four perfect squares (1, 9, 81 and 729), so 4 is the mathematically correct answer.
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