The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # Practice Paper 4 Question 18

A poll with 2 choices is run among $$n>2$$ participants. Each participant chooses at random. The poll shows the results for the 2 options as percentages rounded to the nearest integer, i.e. $$x$$ rounded is $$\lfloor x+0.5\rfloor$$, where $$\lfloor x\rfloor$$ is the greatest integer less than or equal to $$x$$. For example, 1/3 would be shown as 33%. What is the probability that the sum of the 2 shown percentages does not add to 100?

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## Warm-up Questions

1. Why does $$\lfloor x+0.5\rfloor$$ round $$x$$ to the nearest integer? Use the ceiling function $$\lceil \cdot \rceil$$ to create a rounding function.
2. Under what circumstances is $$\lfloor a+1 \rfloor = \lceil a \rceil$$ true?
3. Boris flips $$8$$ fair coins. What is the probability that at least $$6$$ of them are tails?

## Hints

• Hint 1
How are the two percentages related?
• Hint 2
Under what circumstances do the two rounded percentages not add up to 100? Try expressing it in an equation with integer solutions.
• Hint 3
The only scenarios in which the rounded percentages do not add up to 100 are when one of the percentages end in $$.5,$$ causing both percentages to round up and their sum to be 101.
• Hint 4
Try phrasing the above condition in terms of $$n, k$$ and $$j,$$ where $$k$$ is the number of votes for the first option and $$j$$ is the integer of that percentage.
• Hint 5
Why not rearrange to get rid of all fractions and decimal points?
• Hint 6
Consider the prime factors of the terms in your equation. What can you deduce about the divisibility of $$n?$$
• Hint 7
$$n$$ must be divisible by $$8.$$ By substituting $$n = 8m,$$ could you derive a new equation in terms of $$n, k$$ and $$j?$$
• Hint 8
What are the possible factors of $$m?$$ How about doing a case split?
• Hint 9
By considering divisibility of $$5,$$ determine which values of $$j$$ are possible.
• Hint 10
Given the possible values for $$j,$$ what are the possible values for $$k?$$
• Hint 11
What is the probability that the $$n$$ voters will choose one of the suitable $$k?$$

## Solution

It suffices to consider the answers picking one of the options; suppose there are $$k$$ of these. Then the rounded percentages of the two answers are $$\lfloor \frac{100k}{n} + 0.5\rfloor$$ and $$\lfloor \frac{100(n-k)}{n} + 0.5\rfloor$$ respectively. Adding these together we get $$100 + \lfloor \frac{100k}{n} + 0.5 \rfloor$$ $$+ \lfloor 0.5 - \frac{100k}{n}\rfloor,$$ which evaluates to 101 when the fractional part of $$\frac{100k}{n}$$ is exactly $$0.5.$$

Thus, we need to solve $$\frac{100k}{n} = j + 0.5$$ for $$j \in \{0, \dots, 99\}.$$ Rearrange to obtain $$200k = n(2j+1).$$ Since $$200 = 2^3 \cdot 5^2$$ and $$2j+1$$ must be odd, it must be the case that $$n$$ is a multiple of $$2^3 = 8.$$ Writing $$n=8m,$$ our expression becomes $$5^2k=m(2j+1)$$ for $$j\in \{0, \ldots, 99\}.$$ There are three cases:

1. $$m$$ is not a multiple of $$5.$$ This means that $$2j+1$$ must be a multiple of $$25,$$ hence the only possible values for $$2j+1$$ are $$S_1 = \{25, 75, 125, 175\}.$$
2. $$m$$ is a multiple of $$5$$ but not a multiple of $$25.$$ This means that $$2j+1$$ must be a multiple of $$5,$$ hence the only possible values for $$2j+1$$ are $$S_2 = \{5,15,25,35\ldots,195\}.$$
3. $$m$$ is a multiple of $$25.$$ There is no restriction on $$j$$ except $$j<100,$$ hence the only possible values for $$2j+1$$ are $$S_3 = \{1,3,5,7\ldots,199\}.$$

Each $$x \in S_i,$$ $$K(x) = \frac{nx}{200}$$ will result in the rounded percentages to not add up to $$100.$$ Thus, the probability is:

$\begin{equation*} p(n)= \begin{cases} 0 & \text{if } 8\nmid n \\ 2^{-n}\sum_{x \in S_1}\binom{n}{K(x)} & \text{if } 8\mid n, 5\nmid n \\ 2^{-n}\sum_{x \in S_2}\binom{n}{K(x)} & \text{if } 40\mid n, 25\nmid n \\ 2^{-n}\sum_{x \in S_3}\binom{n}{K(x)} & \text{if } 200 \mid n \\ \end{cases} \end{equation*}$

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