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# The cube root of what integer power of 2 is closest to 50?

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Math Expert
Joined: 02 Sep 2009
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The cube root of what integer power of 2 is closest to 50?  [#permalink]

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14 Sep 2015, 04:01
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31
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Difficulty:

95% (hard)

Question Stats:

40% (02:05) correct 60% (02:05) wrong based on 434 sessions

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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.

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Math Expert
Joined: 02 Sep 2009
Posts: 52361
Re: The cube root of what integer power of 2 is closest to 50?  [#permalink]

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20 Sep 2015, 20:26
1
6
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.

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The cube root of what integer power of 2 is closest to 50?  [#permalink]

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14 Sep 2015, 17:46
11
1
$$50^3 = 5^3 * 10^3 = 5^3 * 5^3 * 2^3$$

$$5^3$$ is 125 and I know that $$2^7$$ is 128.

Thus,
$$2^7 * 2^7 * 2^3 = 2^{17}$$

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The cube root of what integer power of 2 is closest to 50?  [#permalink]

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14 Sep 2015, 05:24
2
2
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

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Re: The cube root of what integer power of 2 is closest to 50?  [#permalink]

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14 Sep 2015, 11:18
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.
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Re: The cube root of what integer power of 2 is closest to 50?  [#permalink]

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06 Nov 2015, 09:50
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?  [#permalink]

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06 Nov 2015, 10:33
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?  [#permalink]

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12 Apr 2017, 17:08
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?

$$(2^{17})^\frac{1}{3}=2^\frac{17}{3}=2^{5.6}=50ish$$
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Re: The cube root of what integer power of 2 is closest to 50?  [#permalink]

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19 Apr 2017, 22:09
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=64-32+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?  [#permalink]

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26 Jul 2018, 04:58
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?  [#permalink]

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30 Jul 2018, 18:49
1
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
Re: The cube root of what integer power of 2 is closest to 50? &nbs [#permalink] 30 Jul 2018, 18:49
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