Manhattan GMAT Challenge Problem of the Week - 9 Aug 2011

by Manhattan Prep, Aug 9, 2011

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Question

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

A. 16

B. 17

C. 18

D. 19

E. 20

Answer

The underlying math facts that you need to know for this problem are the powers of 2, through [pmath]2^10[/pmath]. Know these powers, or be able to rederive them quickly.

[pmath]2^1[/pmath] = 2

[pmath]2^2[/pmath] = 4

[pmath]2^3[/pmath] = 8

[pmath]2^4[/pmath] = 16

[pmath]2^5[/pmath] = 32

[pmath]2^6[/pmath] = 64

[pmath]2^7[/pmath] = 128

[pmath]2^8[/pmath] = 256

[pmath]2^9[/pmath] = 512

[pmath]2^10[/pmath] = 1,024

We are asked what integer power of 2 gives us a cube root close to 50. In other words, [pmath]50^3[/pmath] should be close to that power of 2.

In equation form, we have [pmath]2^n[/pmath] [pmath]50^3[/pmath]. What is n?

One approach is to find [pmath]50^3[/pmath], which equals ([pmath]5^3[/pmath] )([pmath]10^3[/pmath] ) = 125 1,000 or 125,000.

Look at our list of powers of 2 up to [pmath]2^10[/pmath], and lets match up to 125,000.

125 is approximately [pmath]2^7[/pmath] (= 128), while 1,000 is approximately [pmath]2^10[/pmath] (= 1,024).

So 125,000 is approximately [pmath]2^7[/pmath] [pmath]2^10[/pmath], or [pmath]2^17[/pmath]. So n = 17. No other power of 2 is even close to 125,000. By the way, [pmath]2^17[/pmath] 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 [pmath]2^3[/pmath]:

[pmath]2^n[/pmath] [pmath]50^3[/pmath]

[pmath]2^3[/pmath] [pmath]2^n[/pmath] [pmath]2^3[/pmath] [pmath]50^3[/pmath]

[pmath]2^(n+3)[/pmath] [pmath]100^3[/pmath] = 1,000,000 = [pmath]1,000^2[/pmath].

Since [pmath]2^10[/pmath] 1,000, we know that [pmath]1,000^2[/pmath] [pmath](2^10)^2[/pmath] = [pmath]2^20[/pmath].

So n + 3 = 20, or n = 17.

The correct answer is B.

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