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The cube root of what integer power of 2 is closest to 50?
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14 Sep 2015, 05:01
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Re: The cube root of what integer power of 2 is closest to 50?
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20 Sep 2015, 21:26
Bunuel wrote: The cube root of what integer power of 2 is closest to 50?
A. 16 B. 17 C. 18 D. 19 E. 20
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:The underlying math facts that you need to know for this problem are the powers of 2, through 2^10. Know these powers, or be able to rederive them quickly. 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 2^9 = 512 2^10 = 1,024 We are asked what integer power of 2 gives us a cube root close to 50. In other words, 50^3 should be close to that power of 2. In equation form, we have 2^n ≈ 50^3. What is n? One approach is to find 50^3, which equals (5^3 )(10^3 ) = 125 × 1,000 or 125,000. Look at our list of powers of 2 up to 2^10, and let’s match up to 125,000. 125 is approximately 2^7 (= 128), while 1,000 is approximately 2^10 (= 1,024). So 125,000 is approximately 2^7 × 2^10, or 2^17. So n = 17. No other power of 2 is even close to 125,000. By the way, 2^17 equals 131,072, but the last thing you should do is calculate that number exactly. You could also take the original equation and multiply in 2^3: 2^n ≈ 50^3 2^3 × 2^n ≈ 2^3 × 50^3 2^(n+3) ≈ 100^3 = 1,000,000 = 1,000^2. Since 2^10 ≈ 1,000, we know that 1,000^2 ≈ (2^10)^2 = 2^20. So n + 3 = 20, or n = 17. The correct answer is B.
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The cube root of what integer power of 2 is closest to 50?
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14 Sep 2015, 18:46
\(50^3 = 5^3 * 10^3 = 5^3 * 5^3 * 2^3\)
\(5^3\) is 125 and I know that \(2^7\) is 128.
Thus, \(2^7 * 2^7 * 2^3 = 2^{17}\)
Answer: B




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The cube root of what integer power of 2 is closest to 50?
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14 Sep 2015, 06:24
Bunuel wrote: The cube root of what integer power of 2 is closest to 50?
A. 16 B. 17 C. 18 D. 19 E. 20
Kudos for a correct solution. Integer powers of 2 = 2 , 2^2, 2^3, 2^4, 2^5, 2^6, 2^7 etc. Let, Cube root of x is closest to 50 i.e. x is closest to cube of 50 = 50*50*50 = 125000 But x must be equivalent to an Integer power of 2 2^10 = 1024 2^7 = 128 i.e. 2^17 = 128*1024 = 128000 (approx.) i.e. 2^18 = 256*1024 = 256000 (approx.) i.e. 2^16 = 64*1024 = 64000 (approx.) i.e. 2^17 which is closest to 125000 i.e. Integer power of 2 must be 7 Answer: option B
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Re: The cube root of what integer power of 2 is closest to 50?
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14 Sep 2015, 12:18
Cube of 50 = 125,000 From the options: 2^16 = 2^10*2^6 = 1024*64 ≈ 64,000 2^17 = 2^10*2^7 = 1024*128 ≈ 128,000 2^16 = 2^10*2^8 = 1024*256 ≈ 256,000 Integer power of 2^17 is closest to 50^3 → Cube root of 2^17 is closest to 50. Answer: B
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Re: The cube root of what integer power of 2 is closest to 50?
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06 Nov 2015, 10:50
Hey guys from the setup I did this fairly quickly with the LOG function. Maybe someone could check it over. I find using logs pretty easy
cubic root of 2^n = 50
2^n=50^3
Log both sides to solve for n
n*log(2) = 3*log(50) n*log(2) = 3*[log(5)+log(10)] 0.301n=3[0.698+1]
*simplied and rounded since they are pretty close to tenths
0.3n=3*(.7+1) 0.3n=2.1+3 0.3n=5.1 n=5.1/(3/10) n=17



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Re: The cube root of what integer power of 2 is closest to 50?
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06 Nov 2015, 11:33
GMATDemiGod wrote: Hey guys from the setup I did this fairly quickly with the LOG function. Maybe someone could check it over. I find using logs pretty easy
cubic root of 2^n = 50
2^n=50^3
Log both sides to solve for n
n*log(2) = 3*log(50) n*log(2) = 3*[log(5)+log(10)] 0.301n=3[0.698+1]
*simplied and rounded since they are pretty close to tenths
0.3n=3*(.7+1) 0.3n=2.1+3 0.3n=5.1 n=5.1/(3/10) n=17 Yes that is an acceptable method but only if you are comfortable with the log function and remeber the values for log 2, log 3 and log 5.



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The cube root of what integer power of 2 is closest to 50?
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12 Apr 2017, 18:08
Bunuel wrote: The cube root of what integer power of 2 is closest to 50?
A. 16 B. 17 C. 18 D. 19 E. 20
Kudos for a correct solution. Hey guys I'm not sure if I'm misunderstanding the question but is my approach totally incorrect? I tested the answer choices: \((2^{17})^\frac{1}{3}=2^\frac{17}{3}=2^{5.6}=50ish\)



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Re: The cube root of what integer power of 2 is closest to 50?
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19 Apr 2017, 23:09
Given: 3√2^n=2^(n/3) ≈ 50
50 is somewhere between 2^5 (which is 32) and 2^6 (which is 64).
If 2^5 > n/3 = 5 > n=15 If 2^6 > n/3 = 6 > n=18
Now, we know that n is somewhere between 15 and 18, which are 3 units apart. 2^5 (=32) and 2^6 (=64) are 33 units apart (33=6432+1). Thus, every 1 unit of n is 11 units (11=33/3) of 2^1k. Or 3:33 > 1:11
Thus, 2^6 (which is 64)  11 = 53 (closest to 50) and n (which is 18 for n^6)  1 = 17.



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Re: The cube root of what integer power of 2 is closest to 50?
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26 Jul 2018, 05:58
Bunuel generis I chose B based on following reasoning. I solved it backwards or sidewards Option A) 16 *3 = 48 Option B) 17*3 = 51 so here I see that 51 is closer to 50 than 48 to 50, hence B
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Re: The cube root of what integer power of 2 is closest to 50?
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30 Jul 2018, 19:49
This is how i solved it: 2^n = 50^3 2^n = (5^2 * 2^1)^3 2^n = 125*125* 2^3 2^n = 2^7 * 2^7 * 2^3 2^n = 2^(7+7+3) 2^n = 2^17 n = 17 Answer is B




Re: The cube root of what integer power of 2 is closest to 50? &nbs
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