2014 AMC 12A Problem 23

2014 AMC 12A 23. The fraction

\displaystyle\frac{1}{99^2} = 0.\overline{b_{n-1}b_{n-2} \dots b_2 b_1 b_0},

where n is the length of the period of the repeating decimal expansion. What is the sum b_0 + b_1 + \dots + b_{n-1}?

(A) 874   (B) 883   (C) 887   (D) 891   (E) 892

Before I start discussing the solution to this post, I must apologise that I haven’t been able to find a good way to represent long division in WordPress.

How can we find the repeated part of the decimal expansion? We can do so by treating 1 as 1.0000 \dots and perform long division, just as we did in primary school.

For example, how do we know that \frac{1}{2} = 0.5 and that \frac{1}{3} = 0.\overline{3}? By long division and noticing the repeated patterns:

AMC_12A_23_01AMC_12A_23_02

Should we try to divide 1.0000 \dots by 9801 (i.e. 99^2) then? It’s possible, but looking at the multiple choice options given, it looks like we will be dividing for a very long time…

(Sidenote: When tackling a multiple choice question, take a look at the options given before starting to solve the problem. Sometimes the options can give you hints and how to solve to problem or which lines of attack may not work.)

While dividing 1.0000 \dots by 9801 doesn’t seem feasible, perhaps dividing by 99 could work? After all, the question does tell us to consider the fraction \frac{1}{99^2} and not \frac{1}{9801}, so \frac{1}{99} should have some significance.

AMC_12A_23_03

As we can see from the working above, \frac{1}{99} = 0.\overline{01}. That seems nice enough! Armed with this fact, how can we obtain the decimal expression for \frac{1}{99^2}?

Whenever we encounter infinity, it’s a good idea to try small cases to get an idea of what the “infinite” case would look like. In this case, small cases would refer to truncating the decimal expansion of 0.\overline{01} to see if we can find a pattern.

Below we have 3 examples of this, where we have multiplied 0.\overline{01} by 0.01, 0.0101 and 0.010101:

AMC_12A_23_04

AMC_12A_23_05

AMC_12A_23_06Looks like we have a pattern! If we were to extrapolate our pattern, the decimal part of 0.\overline{01} \times 0.\overline{01} would be 00, followed by 01, followed by 02, followed by 03, …

There’s just one slight snag: since each step is allotted 2 digits in the decimal expansion, what would it look like when we hit “followed by 100”? Well, if we add it up, we get the following:

AMC_12A_23_07

The red 9 & 1 add up to the red 0 at the bottom and carry a 1 over to the blue 9. This in return results in the blue 0 at the bottom and carries a 1 over to the green 8. This results in a green 9 at the bottom.

Hence, \frac{1}{99^2} = 0.\overline{00010203\dots969799}. It remains to add up the digits of 00, 01, 02, …, 96, 97, 99. One could do it manually, or one could notice that the sum of digits is the sum of all the 2-digit numbers (allowing 0 to be a leading digit) except 98. Each digit appears exactly 10 times in each digit column, and so each digit appears exactly 20 times. Thus,

\begin{aligned} b_0 + b_1 + \dots + b_{n-1} &= 20 \times (0 + 1 + 2 + \dots + 9) - 9 - 8 \\  &= 20 \times 45 - 17 \\  &= 883. \end{aligned}

The answer is (B).

(Sidenote 1: You might argue that the addition (the step with the coloured digits) is not a watertight proof. What happens when we hit “followed by 200”? Or “followed by 1000”? It’s true that the above is not watertight, but remember that the AMC is a multiple choice competition that gives you just 3 min per question. That’s not a whole lot of time!)

(Sidenote 2: OK, say it’s not the AMC. Is there a way to prove this without any handwaving? Well, there doesn’t appear to be one short of doing long division of 9801 into 1.0000… But now that we know the period of the recurring decimal is 198 digits long, this method just requires a bit of perseverance to do long division for 1.000 (with about 200 zeros behind the decimal point)…)

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