Fermi Questions

Fermi Questions was a Division C event from 2005-2008 and 2012-2013, and it has returned at the national level in 2018 and 2019.

What is a Fermi Question?
A Fermi question is one where a seemingly impossible-to-calculate answer is estimated. A famous example of a Fermi question is "How many licks does it take to get to the center of a tootsie roll pop?", where there is very little data to use and assumptions must be made. Fermi questions are named after Enrico Fermi, a physicist who is known for solving these types of questions.

In Science Olympiad, answers to Fermi questions are given in powers of ten. For example, an estimated answer to the above question of 400 licks is put in scientific notation as [math]4\cdot 10^2[/math], and the exponent on the ten is used as the answer, yielding [math]\boxed{2}[/math]. If the estimate was 600 licks, or [math]6\cdot 10^2[/math], then the answer would be [math]\boxed{3}[/math], rounding up.

Points are usually given as follows:
 * 5 points for the correct power of ten
 * 3 points for one away from the correct power of ten
 * 1 point for two away from the correct power of ten.

For example, if the correct answer to the number of licks to the center of a tootsie roll pop is [math]\boxed{2}[/math], and the given answer is [math]\boxed{2}[/math], five points are awarded. If the given answer is [math]\boxed{3}[/math] or [math]\boxed{1}[/math], 3 points are awarded, and if the given answer is [math]\boxed{4}[/math] or [math]\boxed{0}[/math], one point is awarded. All other answers would receive 0 points.

Sample Question
Here is a Fermi question, worked out, with explanations.

How many pieces of paper could a package of pencil lead cover?

Determining the Facts
The first step in solving a Fermi question is to determine what facts are necessary. We change the desired quantity into more and more manageable quantities so we can make good assumptions:

[math]\text{(Number of paper a package covers)}\\ = \text{(Pieces of paper a piece of lead covers)}\cdot\text{(Number of pieces of lead are in a package)} \\ = \text{(Area of paper a piece of lead covers)}/\text{(Area of paper)}\cdot\text{(Number of pieces of lead are in a package)} \\ = \text{(Width of line drawn by lead)(Length of line a lead can draw)}/\text{(Area of paper)}\cdot\text{(Pieces of lead)}. [/math]

It is often a good idea to represent the quantities with symbols, using appropriate subscripts to mitigate confusions:

Let [math]w_l[/math] be the width of the line drawn by pencil lead, [math]l_l[/math] be the total length of the line a pencil lead can draw, [math]A_p[/math] be the area of paper, and [math]n[/math] be the number of pieces of lead in a package. Then, the desired quantity is [math]\frac{w_l\cdot l_l\cdot n}{A_p}.[/math]

Making the Assumptions
The second step is to make assumptions to the unknown information. The key to Fermi questions is having good assumptions. Obviously, the best assumptions are the ones that aren't assumed, like the size of a piece of paper. However, more often than not, assumptions will be necessary.

It is common knowledge that a piece of paper is 8.5 inch by 11 inch, so [math]A_p=8.5\text{ in}\cdot11\text{ in}\approx 100\text{ in}^2[/math] Notice how we were not precise in our calculation - the event does not require a high level of precision!

The number of pieces of lead in a package vary greatly, so it's not very important to be accurate. 10 pieces seems low, and 1000 seems too high. 100 still seems a little high, so we choose [math]n=70[/math] as the number of pieces of lead in a package.

Most lead pieces are of width 0.5mm or 0.7mm. They are relatively close to each other, so we choose [math]w_l=0.5\text{mm}[/math] because it is much easier to work with.

To find [math]l_w[/math], we have to use common sense: It may be 10 meters, but that seems a little short. It could be 1 km, but that seems a little long. 100 m still seems a little short, but close than 1000 m (1 km). Therefore, we will use [math]l_w=300 m[/math], keeping in mind it could still be small.

Doing the Calculations
We now want to plug the quantities into the formula we created to find the answer. However, many of the quantities are in different units! We need to convert them all into the same unit. Here, cm and m are both practical, and we choose to convert everything to cm.

Since [math]1\text{ in}\approx2.5\text{ cm}, 1\text{ m}=100\text{ cm}, 1\text{ mm}=0.1\text{ cm}[/math], we find that [math]A_p\approx 6\cdot10^2\text{ cm}^2, w_l=5\cdot10^{-2}\text{ cm}, l_l=3\cdot10^4\text{ cm}.[/math]

Therefore, the desired quantity is [math]\frac{5\cdot10^{-2}\text{ cm}\cdot3\cdot 10^4\text{ cm}\cdot 70}{6\cdot10^2\text{ cm}^2}=175[/math]. Therefore, our answer is [math]\boxed{2}[/math].

Alternative Method
You may have a good sense of how many pieces of paper a piece of lead can cover. In that case, we let [math]p_l[/math] be the number of paper a piece of lead can cover, and we want to find [math]p_l\cdot n.[/math]

From experience sketching in art classes, a typical detailed sketch takes a third of a pencil lead, so a piece of lead lasts 3 sketches. Since a sketch doesn't cover the entire paper, but leaves about half of it as white space, we can assume [math]p_l=\frac{3}{2}=1.5[/math].

Then, [math]p_l\cdot n=1.5\cdot 70\approx 100[/math], giving us an answer of [math]\boxed{2}[/math].

Some contestants may be extremely confident about the value [math]p_l[/math], while others may have little idea about it. However, while the expression you try to calculate may be different, the process is the same: we change the unknown quantity into ones we are somewhat confident about, we make assumptions, and we calculate the unknown quantity.

Magnitude Notation
The above question was worked out sequentially with numbers in to allow for explanation. This works well for shorter problems, but with longer problems with longer numbers, this can take a while, and can become a problem with many questions and limited time and space. When doing such problems, it is better to note numbers in a format that is easier to work with. This format is called magnitude notation.

Below are two methods for using Magnitude notation: Method 1

For a number in scientific notation [math]a\cdot 10^b[/math], we write it as aEb. Recall that [math]1\le a<10[/math]: This does not have to be strictly followed in intermediate calculations, but you must convert it into this form to get the final answer.

For example, [math]1\cdot 10^3[/math] would be 1E3, [math]6\cdot 10^{23}[/math] would be 6E23, [math]7\cdot 10^{-34}[/math] would be 7E-34.

[math]a[/math] should always have 1 or 2 significant figures. If it has 3 or more, round the number because that level of precision is usually unnecessary given the time constraint.

Multiplication
To multiply two numbers aEb and cEd, we add [math]b[/math] and [math]d[/math] and multiply [math]a[/math] and [math]c[/math]. If [math]a\cdot c\ge10[/math], we divide [math]a\cdot c[/math] by 10 and add 1 to the exponent.

For example,

7E4 * 4E6 =(7*4)E(4+6)=28E10=2.8E11,

2E6 * 3E-9=(2*3)E(6-9)=6E-3,

5E4 * 2E-5=(5*2)E(4-5)=10E-1=1E0.

Round your result if the future multiplications start to become difficult.

Addition and Subtraction
To add/subtract two numbers aEb and cEd, we change the numbers so [math]b[/math] and [math]d[/math] have the same value (usually the larger one), modifying [math]a[/math] and [math]c[/math] along the way. We then add/subtract [math]a[/math] and [math]c[/math].

For example,

2E5 + 4E5 = 6E5,

7E4 + 4E6 = 0.07E6+4E6 = 4.07E6 ≈ 4.1E6,

6E6 + 5E5 = 6E6+0.5E6 = 6.5E6,

9E6 - 9E5 = 9E6-0.9E6 = 8.1E6,

4E5 - 5E6 = 0.4E6-5E6 = -4.6E6,

2E6 + 9E3 = 2E6+0.009E6 ≈ 2E6!

If b and d differ by more than 2, ignore the number with the smaller magnitude because it would be almost negligible.

Method 2


 * [math]1\cdot 10^3[/math] would be E3
 * [math]4\cdot 10^3[/math] would be +E3
 * [math]7\cdot 10^3[/math] would be -E4

Basically, the number is put in exponential notation (like your calculator does) and rounded so there is only one digit.

Then, if the one digit is 0, 1 or 2, leave the exponent part (E3), but do not put a plus or minus. If the digit is an 8 or 9, add one to the exponent and do not put a plus or minus. If the digit is a 3 or 4, then you leave the exponent and put a plus. If the digit is a 6 or 7 add one to the exponent and put a minus. If 5 is rounded up, put add one to the exponent and put a minus. If five is rounded down, then put a plus.

Multiplication
Magnitude notation has a few important rules regarding multiplication.


 * With magnitude notation, when two numbers with plus signs are multiplied, the pluses are removed and one is added to the resulting exponent. for example, +E5 times +E7 equals E13.
 * When two numbers, one with a +, and one with a -, are multiplied, the signs are canceled without changing the exponent.
 * When two numbers, both with minus signs, are multiplied, one is subtracted from the exponent and the signs removed.

With division, same signs cancel. Opposing signs are removed, adding one to the exponent of the value with the + sign.

Addition and Subtraction
Addition and subtraction have somewhat complicated rules.


 * If the exponents are equal:
 * If the signs are both pluses, add one to the exponent and remove the signs.
 * If there are no signs, put a plus (exponents still equal).
 * If both signs are minus, remove the signs (exponents still equal).
 * If one exponent is one number larger than the other:
 * Remove a minus sign if there is one on the larger one
 * Add a plus sign if there is no sign on the larger one
 * Increase the exponent on the larger one while removing the sign if there is a plus sign on the larger one.
 * If the exponents are 2 or more apart, simply ignore the smaller number.

Although magnitude notation appears more complicated, it is faster and neater to work with, and thus easier to use with limited time and space.

Equations
In Fermi Questions, there is little need for accurate complex equations. Therefore, use simplified versions of the needed formulas. If a particular formula is unknown, make up an equation that makes sense. For example, instead of calculating values using a parabola, use the triangle formula instead, slightly increasing the base and height of the triangle to compensate for the curve. Likewise, for an ellipsoid, use a cylinder, decreasing the length to compensate for the curve.

Calculation Problems
In some tournament, there are purely calculation problems with little assumption. For example, one may be asked to evaluate [math]2^{188}[/math], [math]7^{65}[/math] or [math]17!=17\cdot 16\cdot\ldots \cdot2\cdot1[/math]. While many tournaments do not include these problems as they are calculation-heavy, in 2017 it appeared in Clements (JETS) Invitational, Seven Lakes Invitational, Islip Invitational, Princeton Invitational, among many other Invitationals.

For these problems, a bit of advanced math background is necessary:

When reporting an answer [math]A[/math], we are not reporting the answer but rather [math]\log_{10}A[/math], where [math]\log[/math] is the Logarithm Function. For example, [math]200[/math] is reported as [math]\log_{10}200=2.3\approx \boxed{2}[/math], and [math]900[/math] is reported as [math]\log_{10}900=3.0\approx\boxed{3}[/math]. Note, however, that [math]\log_{10} 4.99\approx0.7[/math], so when we find the logarithm of the answer, we round down if the decimal part is less than [math]0.7[/math]. We will denote [math]\log[/math] as logarithm base 10.

For math problems with daunting numbers, it is often easier to work with logarithm of the answer rather than the answer itself using log rules:

Since [math]\log_{10}a^b=b\log_{10}a[/math], [math]\log_{10}2^{188}=188\log_{10}2[/math]. It is famous (and worth remembering!) that [math]\log_{10}2\approx0.3[/math], so the answer is [math]188\cdot0.3\approx\boxed{56}[/math]. Note that we don't need to convert into scientific notation - the logarithm finds the answer directly!

Similarly, [math]\log_{10}7^{65}=65\log_{10}7\approx 65\cdot 0.85\approx\boxed{55}[/math].

To find factorials, denoted by [math]![/math], we either use direct calculation or Stirling's approximation:

If the number before the factorial is small (<11), directly calculate the answer. For example, [math]8!=8\cdot7\cdot\ldots\cdot1=40320[/math], giving us an answer of [math]\boxed{4}[/math]. A list of the first few factorials, including [math]0!=1[/math], can be found here.

For larger factorials, we use Stirling's approximation [math]n!\approx\sqrt{2\pi n}(\frac{n}{e})^n, \log_{10}n!\approx n\log(\frac{n}{e})+\log(\sqrt{2\pi n}).[/math]

For example, [math]\log_{10}17!\approx17\log(\frac{17}{e})+\log(\sqrt{6.28\cdot17})\approx 17\log(6) + \log(10)\approx 16\cdot 0.78+1\approx\boxed{14}[/math].

The equation can be further simplified to (n times log n) minus (n times log e) plus log(square root of 2n times pi) where log e equals .4342944 in order to not divide by e and be more precise

Using logarithms, however, require higher precision: you need precision to the tenth place to get full credit, often meaning 2-3 significant figures. Therefore, it is important to know values of [math]\pi\approx3.14, e\approx2.72[/math] to 2-3 significant figures and be able to multiply/divide numbers with 2-3 significant figures quickly.

It is also important to know the logarithm of 2-9: [math]\log(2)\approx0.3, \log(3)\approx 0.48, \log(4)\approx 0.6, \log(5)\approx0.7, \log(6)\approx 0.78, \log(7)\approx 0.85, \log(8)\approx0.9, \log(9)\approx 0.95.[/math]. You can derive the rest using the formula [math]\log_{10}(a\cdot 10^b)=b+\log_{10}(a).[/math]

Competition Tips
Teams can do very well in Fermi Questions by being as accurate as possible with the most questions. This can be done in certain ways.

Partner Pairing
In Fermi questions, the partner pairing is important. You generally want two people with different interests and personalities. The ideal pair would be the estimator and the number cruncher.

Estimator
The estimator should be a visual/kinetic (but more visual) learner with a good memory. They need to be able to estimate the dimensions of, for example, a football stadium based on their memory of one. They need to be able to estimate the weights of objects they know the visual size of, but may or may not have held before. He (or she) should also know random facts, like the frequency of a cordless phone, or the number of Crayola colors. But above all, the estimator must be able to determine all of these estimations with relative accuracy.

Number Cruncher
The number cruncher is someone quick to perform the calculations. The number cruncher should know physical values and conversion factors, like how many pounds are in a kilogram, the speed of light, speed of sound, etc.

The number cruncher is less crucial to event success. The estimator can memorize the values, but the number cruncher cannot easily learn how to estimate.

Use Facts
Facts provide a good basis for estimations, with or without an estimator. Because even an estimator is not infallible, it is crucial to know unit conversions, as well as many basic facts about things ranging from stellar distances to volumes of the earth's oceans. Learn esoteric units, like BTU's, barns, outhouses, sheds, furlongs, and others, which helps when answers are requested to be in terms of an unusual unit. There have been questions were students were asked to put answers in terms of a poronkusema, so nothing is off limits.

Useful Facts

 * Physics facts, like the speed of sound, or the wavelength of a certain color of light.
 * Human body facts.
 * State facts from the state in which the competition is taking place
 * Facts about the U.S.A.
 * Facts about the world.
 * Esoteric units, such as those listed above.
 * Infomation about the earth, and the sun (Radius, mass etc)
 * Physics equations and their constants (Gravitational attraction, Escape Velocity etc)

Memorize whatever you can. Even if the fact seems useless, anything can be tested, and a fact can be used to estimate another fact.

Sometimes fermi questions simply ask for an obscure fact, such as the mass of a swallow and the mass of a coconut. As it is impossible to know everything, just guess. Answers are in powers of ten, and therefore, they can be off by a lot and still be right.

Other Tips

 * You can be off by as much as nearly 100,000% and still get points. In most cases, being within 100-200% will earn full points, and being within 1,000% will guarantee at least 3 points.
 * Guess away. There are no penalty for being wrong. Just don't guess blindly. Don't say 3 elephants weigh as much as the titanic.
 * When you round something down, round something else up (or down if you're dividing).
 * These are exponents. E3 plus +E7 is +E7 (this is magnitude notation)
 * The competition is timed. Don't dwell on a single problem unless it's your last.
 * Keep your calculations neat and simple. Often times, space is limited.
 * Use common sense.
 * And most of all, have fun.