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

Expert replies
by Max@Math Revolution » Wed Dec 12, 2018 1:18 am

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Answers

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B

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[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
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Source: — Problem Solving |

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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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by Brent@GMATPrepNow » Wed Dec 12, 2018 8:21 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

How many perfect squares lie FROM 2^4 TO 2^8, inclusive?

A. 12
B. 13
C. 15
D. 18
E. 20
First rewrite 2^4 and 2^8 as SQUARES of integers (aka perfect squares)
2^4 = (2²)² = 4²
2^8 = (2^4)² = 16²

So, we want to find the number of perfect squares FROM 4² to 16² inclusive
Let's list them: 4², 5², 6², 7², 8², 9², 10², 11², 12², 13², 14², 15², and 16²
There are 13 such perfect squares

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Wed Dec 12, 2018 9:28 am
$$2^4=16$$
$$2^8=256$$
$$\text{So, } 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \text{ and } 256$$
$$\text{Total } 13 \text{ perfect squares}$$
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by Max@Math Revolution » Thu Dec 13, 2018 11:44 pm
=>

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
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by Scott@TargetTestPrep » Thu Mar 21, 2019 4:43 pm
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 = 16 and 2^8 = 256.

Since 2^4 = 4^2 and since 2^8 = 16^2, we see that there are 16 - 4 + 1 = 13 perfect squares lie between 2^4 and 2^8, inclusive.

If this is not clear, notice that the perfect squares between 4^2 and 16^2 must be: 5^2, 6^2, 7^2, ... , 13^2, 14^2, and 15^2, for a total of 13 perfect squares.

Answer: B

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