Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

How many perfect squares lie between 2^4 and 2^8, inclusive?

A. 12
B. 13
C. 15
D. 18
E. 20

$${2^4} \le {M^2} \le {2^8}\,\,,\,\,\,M\,\,{\mathop{\rm int}} \,\,\,\mathop \ge \limits^{{\rm{WLOG}}} \,\,\,0\,\,\,\,\,\,\,\,\left( * \right)\,\,\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{WLOG}}\,\,{\rm{ = }}\,\,{\rm{without}}\,\,{\rm{loss}}\,\,{\rm{of}}\,\,{\rm{generality}}\,} \right]$$
$$?\,\,\,:\,\,\,\# \,\,\,{\rm{of}}\,\,{\rm{possibilities}}\,\,{\rm{for}}\,\,\left( * \right)$$
$$\sqrt {{2^4}} \le \sqrt {{M^2}} \le \sqrt {{2^8}} \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,4 = {2^2}\,\,\,\mathop \le \limits^{M\,\, \ge \,\,0} M\,\,\, \le {2^4} = 16\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{fingers}}\,\,{\rm{technique}}} \,\,\,\,\,\,\,? = 16 - 4 + 1 = 13$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

$$2^4=16$$
$$2^8=256$$
$$\text{So, } 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \text{ and } 256$$
$$\text{Total } 13 \text{ perfect squares}$$

=>

We need to count the number of integers n satisfying 2^4 ≤ n^2 ≤ 2^8 or 2^2 ≤ n ≤ 2^4.
The number of integers satisfying 2^2 ≤ n ≤ 2^4 is 2^4 - 2^2 + 1 = 16 - 4 + 1 = 13.

Therefore, the answer is B.
Answer: B