Manhattan GMAT Challenge Problem of the Week – 21 Feb 2011

by Manhattan Prep, Feb 21, 2011

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Question

What is the tenth digit to the right of the decimal point, in the decimal expansion of [pmath]1/5^10[/pmath]?

(A) 0

(B) 2

(C) 4

(D) 6

(E) 8

Answer

There are several ways to attack this problem. One way is to rewrite [pmath]1/5^10[/pmath] as [pmath](1/5)^10[/pmath], and then convert to decimals:

[pmath]1/5^10[/pmath] = [pmath](1/5)^10[/pmath] = [pmath](0.2)^10[/pmath]

Now rewrite 0.2 as 2 [pmath]10^-1[/pmath]:

[pmath](0.2)^10[/pmath] = [pmath](2 * 10^{-1})^10[/pmath] = [pmath]2^10[/pmath] [pmath]10^-10[/pmath]

Multiplying an integer (such as [pmath]2^10[/pmath]) by [pmath]10^-10[/pmath] moves the decimal 10 places to the left. So the digit ten places to the right in the decimal expansion of [pmath]1/5^10[/pmath] is the units digit of the original integer.

If you dont know that [pmath]2^10[/pmath] = 1,024 off the top of your head, you can either find it manually or simply examine the pattern in the units digits of the powers of 2. These units digits repeat themselves:

[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

etc.

The cycle is 2, 4, 8, 6 repeating. After two cycles of four digits, we need two more digits in the cycle, so we arrive at 4.

The correct answer is C.

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