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Alignments to Content Standards: 7.RP.A.2

Task

Coffee costs \$18.96 for 3 pounds.

1. What is the cost for one pound of coffee?

2. At this store, the price for a pound of coffee is the same no matter how many pounds you buy. Let $x$ be the number of pounds of coffee and $y$ be the total cost of $x$ pounds. Draw a graph of the relationship between the number of pounds of coffee and the total cost.

3. Where can you see the cost per pound of coffee in the graph? What is it?

IM Commentary

The purpose of this task is for students to find a unit rate in a context where two quantities are in a proportional relationship and to draw the graph of that proportional relationship.

This is a task where it would be appropriate for students to use technology such as a graphing calculator or GeoGebra, making it a good candidate for students to engage in MP5, Use appropriate tools strategically. A variant of this problem is appropriate for 8th grade; see 8.EE.5 Coffee by the Pound.

Solution

1. You can find the cost for one pound of coffee by dividing the total cost by 3. Coffee costs \$6.32 per pound.

2. We may graph the proportional relationship between the total cost and the number of pounds by plotting the line through the origin and (3, 18.96).

3. The cost of one pound, \$6.32, may be seen on the graph in two ways:

As the point (1, 6.32)

As the slope of the line: \$6.32 per pound.

Note: Students aren't explicitly required to see the connection between the unit rate and the slope until 8th grade (see 8.EE.5) but they may still see it in 7th grade.

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Alignments to Content Standards: 7.RP.A.2 Coffee costs \$18.96 for 3 pounds. What is the cost for one pound of coffee? At this store, the price for a pound of coffee is the same no matter how many pounds you buy. Let $x$ be the number of pounds of coffee and $y$ be the total cost of $x$ pounds.

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Alignments to Content Standards: 8.NS.A

Task

1. Different cultures through history have used many different rational numbers to approximate $\pi$. For each of the historical approximations below, use a calculator to determine approximately how far the fraction is from $\pi$:

1. $\frac{22}{7}\,$

2. $\frac{25}{8}\,$

3. $\left(\frac{9785}{5568}\right)^2\,$

4. $\frac{355}{113}\,$

2. The fraction with denominator 1 which is closest to $\pi$ is $\frac{3}{1} = 3$. For each of the following whole numbers, find the fraction with this denominator which is closest to $\pi$:

3. Look for patterns in the data you've collected thus far. What effect does increasing the denominator have on the accuracy of the best approximation?

IM Commentary

The goal of this task is to explore some important aspects of approximating an irrational number with rational numbers. The irrational number chosen here is $\pi$ because it is one of the most interesting, well known, and grade appropriate irrational numbers. The fact that finite decimal representations are approximations by fractions, whose denominators are powers of $10$, is emphasized by parts ii, iv and vi of the second part of the task.

There is a rich and interesting history of finding approximations of $\pi$. Many details, including the historical approximations in the first part of the question, can be found at the Wikipedia page for $\pi$.

Some of the work done before calculators is truly astounding: Daniel Ferguson found 620 digits of $\pi$ in 1946 without the aid of a calculator. Earlier, in 1873, a British mathematician named Shanks calculated 707 digits but after the 528th place, the calculations were incorrect! For this task, students are highly encouraged to use a calculator. In doing so, it is important to realize that the $\pi$ button on a calculator only gives a (very good) approximation of $\pi$, not an exact value.

The theory of approximating irrational numbers with rational numbers is a rich and old one. The fact that $\pi$ can be extremely well approximated by rational numbers whose denominators are not that large is closely related to the fact that it is an irrational number. It was not known until the second half of the 19th century that $\pi$ is an irrational number and there is still no elementary proof of this fact today. Many of the historical estimates for the size of $\pi$ came from inscribing polygons inside a circle (or circumscribing a polygon around a circle) and then calculating the perimeter of those polygons: this technique, appropriate for high school geometry, is explored in https://www.illustrativemathematics.org/content-standards/tasks/1567.

The task can be extended by having students compare the errors of the approximations with sizes of unit fractions. Particularly good approximations have a small ratio of error to the size of unit fraction. Some such approximations are given by continued fraction convergents, which include the Archimedean and Chinese approximations. Continued fractions have not been part of popular mathematics culture for some time but the theory is fascinating and some information is available here: http://en.wikipedia.org/wiki/Continued_fraction.

Solution

1. We have $\frac{22}{7} \approx 3.142857$ and $\pi \approx 3.14159265$. These differ by a little more than $\frac{1}{1000}$.

2. Since $\frac{25}{8} = 3.125$ we can see that this approximation is not as good as $\frac{22}{7}$. It differs from $\pi$ by about $\frac{17}{1000}$.

3. This approximation is certainly the most complex of the group. Using a calculator we find that $\left(\frac{9785}{5568}\right)^2 \approx 3.088$ and this differs from $\pi$ by a little more than $\frac{1}{20}$. It is the least accurate of all of these estimates.

4. We have $\frac{355}{113} \approx 3.1415929$ which is extremely close to $\pi$. It differs by a few ten-millionths!

1. With denominator 3, the two numbers closest to $\pi$ are $3 = \frac{9}{3}$ and $\frac{10}{3}$. The closer of these is $3$, differing by a little more than 14 hundredths.

2. We have $3.1 \lt \pi \lt 3.2$ and $\pi$ is a little closer to 3.1 than to 3.2.

3. With denominator 20, the closest fractions to $\pi$ are $\frac{62}{20} = 3.1$ and $\frac{63}{20} = 3.15$. The closest of these to $\pi$ is $\frac{63}{20}$.

4. With denominator 100, we see that $\frac{314}{100} \lt \pi \lt \frac{315}{20}$ and $\frac{315}{100}$ is closer. This is the same approximation as $\frac{63}{20}$.

5. In part iii of the first part, we saw that $\frac{355}{113} \approx 3.1415929$ is very close to $\pi$. It is a little bigger than $\pi$ so $\pi$ falls between $\frac{355}{113}$ and $\frac{354}{113} \approx 3.133$. So $\pi$ is closest to $\frac{355}{113}$.

6. With denominator 10000, $\pi$ lies between $\frac{31415}{10000}$ and $\frac{31416}{10000}$. It is closer to $\frac{31416}{10000}$.

3. One trend we can see from the answers to parts (ii), (iv), and (vi) to b is that if the denominator of the fraction is a power of 10 then we can find the closest fraction to $\pi$ by studying its decimal expansion. For example, since $\pi \approx 3.14159265$, we can see both that it lies between 3.1 and 3.2 and that it is closer to 3.1. We also gain an additional decimal of accuracy each time we increase the power of 10. This reasoning only applies when the denominator is a power of ten.

Generally speaking, one would expect larger denominators to help get a better approximation for $\pi$, since there is a smaller distance between successive fractions with that denominator. For example, since fractions of the form $\frac{n}{17}$ (with $n$ an integer) are $\frac{1}{17}$ apart, and since $\pi$ lies between two such numbers, the farthest it could be from the closest one is $\frac{1}{17}$ (in fact, half of this distance, $\frac{1}{34}$).

On the other hand, it is not always true that increasing the denominator permits a more accurate approximation of $\pi$. For example, as calculated above, the denominator 113 allows for a much more accurate estimate than a denominator of even 10000. One thing we can say is that multiplying the denominator by another integer permits at least as good an approximation -- for example, we can get as least as good an approximation with denominator 51 as with 17, for the simple reason that every number of the form $\frac{n}{17}$ can also be written as $\frac{3n}{51}$.

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Alignments to Content Standards: 8.NS.A Different cultures through history have used many different rational numbers to approximate $\pi$. For each of the historical approximations below, use a calculator to determine approximately how far the fraction is from $\pi$: $\frac{22}{7}\,$ $\frac{25}{8}\,$ $\left(\frac{9785}{5568}\right)^2\,$ $\frac{355}{113}\,$ The fraction with denominator 1 which is closest to $\pi$ is $\frac{3}{1} = 3$.

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Alignments to Content Standards: 4.OA.A

Task

There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. Which one grew more?

IM Commentary

The purpose of this task is to foster a classroom discussion that will highlight the difference between multiplicative and additive reasoning. Some students will argue that they grew the same amount (an example of "additive thinking"). Students who are studying multiplicative comparison problems might argue that Jewel grew more since it grew more with respect to its original length (an example of "multiplicative thinking"). This would set the stage for a comparison of the two perspectives. In the case were the students don't bring up both arguments, the teacher can introduce the missing perspective.

In later grades, students will learn that "which grows more" means "which has the greater absolute increase?" and "which has the greater growth rate?" means "which has the greater increase relative to the starting amount?" but students won't see this type of language for two or three years. Teachers need to be aware of this and work to ask questions as unambiguously as possible; for example, when asking for multiplicative comparisons, use language such as, "How many times greater is $x$ than $y$." They should also be prepared to address this potential for confusion along the way.

Solution

Viewing this additively, both snakes grew 2 feet and therefore grew the same amount. Viewing it multiplicatively, Jewel grew $\frac{2}{6}$ its length, while Clyde grew $\frac{2}{8}$ its length. From this perspective, Jewel grew more. Given the purposeful phrasing of the problem, both interpretations are reasonable, but the goal is to understand the two perspectives, thus the difference between additive and multiplicative reasoning.

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Alignments to Content Standards: 4.OA.A There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. Which one grew more?

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Alignments to Content Standards: 8.NS.A

Task

Lucia's calculator gives a value of $$ 3.14159265 $$ for $\pi$.

1. Is the equation $$ \pi = 3.14159265 $$ valid? Explain.

2. When Lucia computes $\pi^2$ on her calculator, using the $\pi$ and square buttons, it shows $$ 9.86960440. $$ On the other hand, when she calculates $(3.14159265)^2$ on her calculator it shows $$ 9.86960438. $$ Explain why the calculator shows different answers for what appears to be the same quantity.

IM Commentary

Calculators are tremendously useful tools but sometimes they can mislead. Whenever dealing with an irrational number, such as $\pi$, the calculator makes an appropriate approximation. For practical purposes this is sufficient. It is important, however, when performing arithmetic operations, to be careful about the effects of working with rounded numbers. The approximation $3.14159265$ for $\pi$ is a little too small, by a little more than three billionths. On the other hand, when asked to find $\pi^2$ the calculator has a built in algorithm which finds the correct answer to the nearest billionth. Calculators provide many opportunities, when appropriately studied, for understanding the mathematics behind its calculations.

This task is intended for instructional (rather than assessment) purposes, providing an opportunity to discuss technology as it relates to irrational numbers and calculations in general. The task gives a concrete example where rounding and then multiplying does not yield the same answer as multiplying and then rounding. The teacher may wish to remind students of the mathematical symbols usually used where Lucia used "=." The symbols $\sim$ and $\approx$ are both used to mean "is approximately equal to" so Lucia could write $\pi \approx 3.14159265$ and this would be correct.

There are many variants on this theme if the teacher wishes to work with an expression other than $\pi$. For example, a calculator will find, to eight decimal places, that $$ \sqrt{2} \approx 1.41421356. $$ If the student copies this number down and then enters it on the calculator and squares it, the calculator will show $1.99999999$ rather than $2$. This also gives some insight into how a calculator works because if the expression on the screen, after calculating $\sqrt{2}$, is squared directly, the calculator will show $2$: in other words, the calculator has stored in its memory, in this case, more information than is shown on the screen!

The teacher may also wish to discuss with students the fact that this issue comes up with fractions as well. If the students compute $\frac{1}{17}$ on a calculator they will find $0.05882353$ (or perhaps a few extra digits) but this can not be correct because the decimal expansion of $\frac{1}{17}$ is not terminating.

Solution

1. The equation $$ \pi = 3.14159265 $$ is not valid. The number $\pi$ is irrational while $3.14159265$ is a rational number. As an irrational number, the decimal expansion of $\pi$ never repeats and $3.14159265$ is the best approximation with $8$ decimal places. To thirty places, the best approximation of $\pi$ is $$ 3.141592653589793238462643383279. $$

For numbers like $\pi$, $e$, and $\sqrt{2}$ which have explicit symbols to represent them, it is best to use the symbol rather than converting to a rounded off decimal. In other words, Lucia can just leave the $\pi$ without trying to convert to a decimal. In computational contexts where what is sought is not an exact expression but an approximate answer, then it would be appropriate and even necessary to write $$ \pi \approx 3.14159265 $$ using the approximately equal to sign $\approx$ instead of the equals sign $=$.

2. When asked to perform the calculation of $(3.14159265)^2$ the calculator will find this quantity and round to the nearest hundred millionth since the display only allows to show the answer to the nearest hundred millionth. On the other hand, when asked to find $\pi^2$, rather than taking an approximate value for $\pi$ and then squaring, the calculator has a built in algorithm which can find the quantity $\pi^2$ to the desired level of precision, in this case to the nearest hundred millionth.

The fact that the calculator reads different numbers for $\pi^2$ and for $(3.14159265)^2$ indirectly answers (a): since these quantities are not equal, it cannot be that $\pi = 3.14159265$.

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Alignments to Content Standards: 8.NS.A Lucia's calculator gives a value of $$ 3.14159265 $$ for $\pi$. Is the equation $$ \pi = 3.14159265 $$ valid? Explain. When Lucia computes $\pi^2$ on her calculator, using the $\pi$ and square buttons, it shows $$ 9.86960440.

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Alignments to Content Standards: 8.EE.A.3

Task

An ant has a mass of approximately $4 \times 10^{−3}$ grams and an elephant has a mass of approximately 8 metric tons.

1. How many ants does it take to have the same mass as an elephant?

2. An ant is $10^{−1}$ cm long. If you put all these ants from your answer to part (a) in a line (front to back), how long would the line be? Find two cities in the United States that are a similar distance apart to illustrate this length.

Note: 1 kg = 1000 grams, 1 metric ton = 1000 kg, 1m = 100 cm, 1km = 1000 m

IM Commentary

In this problem students are comparing a very small quantity with a very large quantity using the metric system. The metric system is especially convenient when comparing measurements using scientific notations since different units within the system are related by powers of ten.

Solution

1. First we observe that the mass of the ant is in grams, where the mass of the elephant is in metric tons. We cannot compare the two masses as is, so we first convert them into the same units. We can do this with ratios, using the conversion chart above to convert metric tons into kilograms and then kilograms into grams.

$$ \begin{align} (8 \text{ metric tons}) \left( \frac{1000 \text{ kg}}{1 \text{ metric ton}} \right) \left( \frac{1000 \text{ grams}}{1 \text{ kg}} \right) &= 8,000,000 \text{ grams}\\ &= 8 \times 10^6 \text{ grams} \end{align} $$

Notice that both metric tons and kg appear once on both the top and the bottom, enabling them to be canceled out, leaving only grams. Also note that we could have converted .003 grams into metric tons, and though this would have given an answer with a large number of decimal places, we would end up with the same final result.

Now that both quantities are in grams, we want to find how many ants, $n$, are required to have the same mass as the elephant. As each ant has a mass of $4 \times 10^{−3}$ grams, we multiply $4 \times 10^{−3}$ by $n$ to get the total mass of the ants $$ \text{total mass of } n \text{ ants} = 4 \times 10^{-3}n $$

and then we set this equal to the mass of the elephant: $$ 4 \times 10^{-3} n = 8 \times 10^6 $$

We find that $n = 2\times10^9$ ants.

Thus, it takes approximately $2 \times 10^9$ ants to have the same mass as an elephant.

2. Each ant is $10^{−1}$ cm long, and we have $2 \times 10^9$ ants total. If we put all these ants in a line (front to back), we can find the length of the line by multiplying the length of one ant by how many ants we have: $$ 2 \times 10^9 \times 10^{−1} \text{ cm} = 2 \times 10^8 \text{ cm} $$

As is, this answer is difficult to illustrate because it is such a large number. To find two objects that are this distance apart, we first want to convert centimeters into more useful units. We do this by using ratios from the above chart, converting centimeters to meters and meters to kilometers. $$ (2 \times 10^8 \text{ cm}) \left( \frac{1 \text{ m}}{10^2 \text{ cm}} \right) \left( \frac{1 \text{ km}}{10^3 \text{ m}} \right) = 2 \times 10^3 \text{ km} = 2000 \text{ km} $$

This amount is much easier to visualize, and it turns out that 2000 km is approximately the driving distance from San Francisco, California to Denver, Colorado.

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Alignments to Content Standards: 8.EE.A.3 An ant has a mass of approximately $4 \times 10^{−3}$ grams and an elephant has a mass of approximately 8 metric tons. How many ants does it take to have the same mass as an elephant? An ant is $10^{−1}$ cm long.

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IM Commentary

This task can be made more hands-on by asking the students to determine about how many pennies are needed to make a stack one inch high. In addition, for part (b) students could be invited to research this question. Useful information which would need to be compiled to get total mintage figures can be found at Annual mintage of Lincoln pennies. From 1909 through 2009 the total number of pennies minted is 455,627,740,918 according to Total mintage of Lincoln pennies.

The goal of this task is to give students a context to investigate large numbers and measurements. Students need to fluently convert units with very large numbers in order to successfully complete this task. The total number of pennies minted either in a single year or for the last century is phenomenally large and difficult to grasp. One way to assess how large this number is would be to consider how far all of these pennies would reach if we were able to stack them one on top of another: this is another phenomenally large number but just how large may well come as a surprise. In particular, before they start working on the problem, the teacher may wish to ask the students whether or not they think it would be possible to reach the moon with this giant stack of pennies.

Four solutions are offered to the problem stressing different mathematical ideas:

1. A division approach taking the distance from the earth to the moon and dividing by the height of a single penny.

2. An algebraic approach where $x$ denotes the number of pennies necessary to reach the moon.

3. A ratio approach which could be solved algebraically, linking to solution (a), or through a ratio table.

4. Arithmetic with scientific notation.

The first of these solutions uses multiplication as well as division because there is conversion of units involved. Note that the second of these solutions leads naturally to the introduction of a function (representing the height of a stack of $x$ pennies) and so, although it also strongly stresses ratio language, it would need to be adapted in order to be appropriate at the sixth grade level. As written, the answer to part (c) meets the 6-RP.3 standard. Students should eventually be comfortable with all three approaches. Note too that throughout the teacher needs to provide some guidance in terms of the level of accuracy with which results should be recorded: because the number of pennies minted is an estimate and the thickness of the pennies is an estimate, no more than one or two significant digits should be recorded in the answers; this is important because it makes each successive calculation easier if numbers are rounded.

The number of pennies made in the last 100 years is about 100 times the number of pennies made in 2011. The vast majority of these pennies, however, have been made in the last 40 years. In 1922, for example, only a little over 7 million pennies were made while in 1982 the total exceeded a staggering 17 billion.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This task helps illustrate Mathematical Practice Standard 4, Model with mathematics. There are 3 parts to this task and each part provides scaffolding opportunities for students as they apply the mathematics they know to find a solution pathway. Because this task has multiple entry points, students may apply several different approaches depending upon their own understanding, such as, arithmetical, algebraic, ratio reasoning, or scientific notation. Students may make assumptions and approximations while solving this task as the number of pennies minted and the thickness of each penny are estimates. As a result, the level of accuracy of results must be resolved which connects to Mathematical Practice 6, Attend to precision.

Solutions

Solution: 1 Arithmetic (6-NS.2 and 6-NS.3)

1. Five billion pennies, each $\frac{1}{16}$ of an inch thick, will make a stack $$ 5,000,000,000 \times \frac{1}{16} \,\,\text{inches} $$ high. Dividing $5,000,000,000$ by $16$ gives a little more than $300,000,000$ inches. We need to convert this to miles so we can begin by finding how many feet there are in $300,000,000$ inches. This will be $$ 300,000,000 \div 12 =25,000,000. $$ Finally, to find out how many miles there are in $25,000,000$ feet we need to divide by $5280$ because there are $5280$ feet in each mile: $$ 25,000,000 \div 5280 \approx 5000. $$ So the stack of pennies would be about $5000$ miles high.

2. A stack of $500$ billion pennies is the same as $100$ stacks of $5$ billion pennies so using the answer from part (a), the stack of $500$ billion pennies would be about $100 \times 5000 = 500,000$ miles high.

3. Here we need to perform the division $239,000 \,\,miles \div \frac{1}{16} inches$. Because the units, miles and inches, are different we need to find out how many inches are in each mile (or equivalently, convert the $239,000$ miles to inches). We have

\begin{align} 239,000\,\,\, \text{miles} & = 239,000 \,\, \text{miles} \times \frac{5280 \,\,\text{feet}}{\text{mile}} \times \frac{12 \,\, \text{inches}}{\text{foot}} \\ &={239,000 \times 5280 \times 12} \,\, \text{inches} \end{align}

Multiplying this out gives $15,143,040,000$ or about $15$ billion inches. Now we can calculate $$ 15,000,000,000 \div \frac{1}{16} = 240,000,000,000. $$ So it takes about $240$ billion pennies to make a stack high enough to reach the moon.

Solution: 2 Algebra (6-EE.6)

1. Since it takes $16$ pennies to make a stack one inch high, with $5,000,000,000$ pennies we could make a stack $$ \frac{5,000,000,000}{16} \text{inches} $$ high. The question asks how many miles this is so we need to convert inches to miles. For this, we use the facts that there are $12$ inches in a foot and $5280$ feet in a mile:

\begin{align} \frac{5,000,000,000}{16} \text{inches} & = \frac{5,000,000,000}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5,000,000,000}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx 5000 \,\, \text{miles}. \end{align}

Note that the exact value of the quotient above is a little more than $4932$ miles but because $5,000,000,000$ is an estimate for the number the number of pennies minted in $2011$ it is apprpriate to list the height of the stack as about $5000$ miles.

2. A stack of $500,000,000,000$ pennies would reach $100$ times as far as a stack of $5,000,000,000$ pennies. So this would be about $100 \times 5000$ miles or about $500,000$ miles.

3. If we write $x$ for the number of pennies in a stack which reaches the moon then we have the equation $$ x \,\times \frac{1}{16} \text{inch} = 239,000 \,\, \text{miles}. $$ We can solve this equation as follows: $$ x \,\,\, \text{inches} = 16 \times 239,000 \,\, \text{miles} \times \frac{5280 \,\,\,\text{feet}}{\text{mile}} \times \frac{12 \,\,\,\text{inches}}{\text{foot}} $$ Simplifying we find

\begin{align} x \,\,\, \text{inches} &= 16 \times 239,000 \times 5280 \times 12 \,\, \text{inches} \end{align}

So $x$ is $16 \times 239,000 \times 5280 \times 12 \approx 240,000,000,000$. So it takes about $240$ billion pennies (or almost a quarter of a trillion!) to reach the moon.

Solution: 3 Ratio (6-RP.3, 8-F.1)

Since the pennies being stacked all have the same thickness, the ratio of the number of pennies in a stack to the height of the stack does not depend on how many pennies are in the stack. In other words if $a$ denotes the number of pennies in a stack and $h$ is the function so that $h(a)$ denotes the height of that stack then the ratios $(a:h(a))$ are equivalent for any positive number $a$ of pennies.

1. Here we have $a = 5,000,000,000$. According to the previous paragraph we need to find $h(5,000,000)$ we need to solve the equation $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( 5,000,000,000: x \,\, \text{miles} \right). $$ We have $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( 5,000,000,000: 5,000,000,000 \times \frac{1}{16}\,\, \text{inches} \right). $$ So to solve for $x$ we need to convert $\frac{5,000,000,000}{16} \text{inches}$ to miles which can be done as in the previous solution:

\begin{align} \frac{5,000,000,000}{16} \text{inches} & = \frac{5,000,000,000}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5,000,000,000}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx 5000 \,\, \text{miles}. \end{align}

So $x$, the height of the stack of $5,000,000,000$ pennies, is about $5000$ miles.

2. If we multiply the number of pennies in a stack by $10$, the height of the stack is also multiplied by $10$ (so that the ratio is the same). Since $h(500,000,000,000) = 10h(50,000,000,000)$ and $h(50,000,000,000)$ is about $5000$ miles, we have that $h(500,000,000,000) \approx 500,000$ miles.

3. We need to find $x$ to solve the following equation of ratios: $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( x: 239,000 \,\, \text{miles} \right). $$ We can first convert miles to inches as in part (a): $$ 239,000 \,\,\text{miles} = 12 \times 5280 \times 239,000 \,\,\frac{\text{inches}}{\text{foot}} \times \frac{\text{feet}}{\text{mile}} \times \text{miles}. $$ Putting this into the previous equation gives $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( x: 12 \times 5280 \times 239,000 \,\, \text{inches} \right). $$ Solving this gives $$ x = 16 \times 12 \times 5280 \times 239,000 \approx 240,000,000,000. $$

Solution: 4 Scientific notation (8-EE.3 and 8-EE.4)

Scientific notation is appropriate for this problem as the number of pennies involved is very large and nicely represented using exponential notation. Either of the above solutions can be adapted for the use of exponential notation and we have chosen here to use the first method.

1. One billion is one with $9$ zeroes or $1 \times 10^9$. So $5$ billion pennies is $5 \times 10^9$ pennies. Since it takes $16$ pennies to make a stack one inch high, with $5 \times 10^9$ pennies we could make a stack $$ \frac{5 \times 10^9}{16} \text{inches} $$ high. The question asks how many miles this is so we need to convert inches to miles. For this, we use the facts that there are $12$ inches in a foot and $5280$ feet in a mile:

\begin{align} \frac{5 \times 10^9}{16} \text{inches} & = \frac{5 \times 10^9}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5 \times 10^9}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx \frac{5 \times 10^9}{1 \times 10^6}\,\, \text{miles}\\ &= 5 \times 10^3 \,\,\text{miles}. \end{align}

In the last step, we use the law of exponents $$ \frac{10^9}{10^6} = 10^{9-6}. $$

2. A stack of $500,000,000,000$ pennies is $5 \times 10^{11}$ pennies or $100$ times as many as the stack of $5 \times 10^9$ considered in part (a). So this stack would reach $100$ times as far as a stack of $5 \times 10^9$ pennies or about $$ 100 \times 5 \times 10^3 = 5 \times 10^5 \,\,\text{miles}. $$

3. If we write $x$ for the number of pennies in a stack which reaches the moon then we have the equation $$ x \,\times \frac{1}{16} \text{inch} = 239,000 \,\, \text{miles}. $$ We can solve this equation as follows: $$ x \,\,\, \text{inches} = 16 \times 239,000 \,\, \text{miles} \times \frac{5280 \,\,\,\text{feet}}{\text{mile}} \times \frac{12 \,\,\,\text{inches}}{\text{foot}} $$ Simplifying we find

\begin{align} x \,\,\, \text{inches} &= 16 \times 239,000 \times 5280 \times 12 \,\, \text{inches} \end{align}

So $x$ is $16 \times 239,000 \times 5280 \times 12 \approx 2.4 \times 10^{11}$. So it takes about $2.4 \times 10^{11}$ pennies (or almost a quarter of a trillion!) to reach the moon.

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This task can be made more hands-on by asking the students to determine about how many pennies are needed to make a stack one inch high. In addition, for part (b) students could be invited to research this question. Useful information which would need to be compiled to get total mintage figures can be found at Annual mintage of Lincoln pennies.

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Alignments to Content Standards: 4.OA.B.4

Task

The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row.

The first student starts with the first locker and goes down the hallway and opens all the lockers.

The second student starts with the second locker and goes down the hallway and shuts every other locker.

The third student stops at every third locker and opens the locker if it is closed or closes the locker if it is open.

The fourth student stops at every fourth locker and opens the locker if it is closed or closes the locker if it is open.

This process continues until all 20 students in the class have passed through the hallway.

1. Which lockers are still open at the end of the game? Explain your reasoning.

2. Which lockers were touched by only two students? Explain your reasoning.

3. Which lockers were touched by only three students? Explain your reasoning.

4. Which lockers were touched the most?

IM Commentary

The purpose of this instructional task is for students to deepen their understanding of factors and multiples of whole numbers. This is a classic mathematical puzzle; often it is stated in terms of 100 lockers, so students have to make the connection to factors and multiples to solve it. In this version, students can just go through all the rounds of opening and closing locker doors and observe after the fact that there is a relationship between the factors a locker number has and whether it is open or closed at the end.

This task provides students with an excellent opportunity to engage in MP7, Look for and make use of structure (if they see early on that there is a relationship with factors and multiples) or MP8, Look for and express regularity in repeated reasoning (if they start to see and describe the pattern as they imagine students opening and closing the lockers).

Because the total number of lockers is only 20, students might answer the questions without thinking about the underlying reasons for their answers. In the first question for example, a student might say "1, 4, 9, 16 are all still open because I tried it out and those were the ones that were left open." If a student goes this route, the teacher can steer the conversation back to factors by asking the students if they notice anything special about the numbers of the lockers which are still open. Asking students to try a larger number of lockers (say 50 or 100) and repeating the game can also help students look for a pattern since going through all the rounds of the game becomes less and less feasible as the number of lockers increases. Once students see the pattern, they should be pressed to explain why it works out this way.

Solution

1. The lockers that are still open are perfect squares, specifically $1\times 1$, $2\times 2$, $3\times 3$, and $4\times 4$. To see why this is true, first note that:

The 1st student changes all the lockers with a locker number that is a multiple of 1.

The 2nd student changes all the lockers with a locker number that is a multiple of 2.

The 3rd student changes all the lockers with a locker number that is a multiple of 3.

The 4th student changes all the lockers with a locker number that is a multiple of 4.

and so on. So a locker is touched by a student if that student's number is a factor of that locker's number.

Now consider the factors of 16 which are 1, 2, 4, 8, and 16. This means the 16th locker was opened by the 1st student, then it was closed by the 2nd student, then it was opened by the 4th student, then it was closed by the second 8th student, and then opened one last time by the 16th student. Now note that the square numbers are the only numbers with an odd number of factors (all other numbers have factors which come in pairs.) Since an odd number of factors means the locker will be changed an odd number of times, it must be open at the end.

With only 20 lockers, students can simply observe this fact about the number of factors to be true for the numbers 1 to 20, which gives practice with finding factors of whole numbers. The explanation for why this is always true goes a bit beyond 4th grade.

2. The lockers that were touched by only two students are the prime numbers (since prime numbers only have two factors.) In our case they are 2, 3, 5, 7, 11, 13, 17, and 19.

3. The lockers which are touched by only three students are the squares of prime numbers (because these are the only numbers which have three factors.) In our case they are 4 and 9. (For example, 4 has three factors: 1, 2, and 4)

4. The lockers which are touched the most are those lockers whose number has the most factors. In our case 12, 18, and 20 all have 6 factors, so they were each touched 6 times. $$ 12 = 1\times 12 = 2\times 6 = 3\times 4$$ $$ 18 = 1\times 18 = 2\times 9 = 3\times 6$$ $$ 20 = 1\times 20 = 2\times 10 = 4\times 5$$

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Alignments to Content Standards: 4.OA.B.4 The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row. The first student starts with the first locker and goes down the hallway and opens all the lockers.