Accurately weighing pennies II

Task

In $1983$ the composition of pennies in the United States was changed due, in part, to the rising cost of copper. Pennies minted after 1983 weigh $2.50$ grams while the earlier copper pennies, from 1865 through 1982, weigh $3.11$ grams. Pennies made between 1859 and 1864 had a different composition, with the same diameter, and weighed $4.67$ grams.

1. A roll of pennies contains $50$ coins. If a roll of pennies weighs $145$ grams (to the nearest hundredth of a gram), how many pennies of each type does the roll contain?
2. If two rolls of pennies have the same weight, do they necessarily contain the same number of pennies of each of the three different weights? Explain.
3. What is the answer to part (b) if the pennies instead weigh $2.5$, $3.1$ and $4.7$ grams respectively?

IM Commentary

This task is a somewhat more complicated version of ''Accurately weighing pennies I'' as a third equation is needed in order to solve part (a) explicitly. Instead, students have to combine the algebraic techniques with some additional problem-solving (numerical reasoning, informed guess-and-check, etc.) Part (b) is new to this task, as with only two types of pennies the weight of the collection determines how many pennies of each type are in the collection. This is no longer the case with three different weights but in this particular case, a collection of 50 is too small to show any ambiguity. This is part of the reason for part (c) of the question where the weight alone no longer determines which type of pennies are in the roll. This shows how important levels of accuracy in measurement are as the answer to part (b) could be different if we were to measure on a scale which is only accurate to the nearest tenth of a gram instead of to the nearest hundredth of a gram.

This task is intended for instructional use only. The awkward numbers and level of difficulty make it unsuitable for assessment purposes.

Solution

1. If we call $x$ the number of pennies of weight $2.50$ grams, $y$ the number of pennies of weight $3.11$ grams, and $z$ the number of pennies of weight $4.67$ grams, then we have $$x + y + z = 50$$ since there are a total of $50$ pennies in the roll. We also have $$2.50x + 3.11y + 4.67z = 145$$ since the total weight of the pennies is $145$ grams. This does not look promising as it is only two equations in three unknowns which is usually not enough information to find values of the variables. We may eliminate $x$ by multiplying the first equation by $2.5$: $$2.5x + 2.5y + 2.5z = 125$$ and then subtract this from the second equation to give $$0.61y + 2.17z = 20.$$ Normally this equation would not be sufficient to solve for $x$ and $y$ but in this case there are severe restrictions on $y$ and $z$: they both have to be non-negative integers and $y + z \leq 50$.

   We can try to solve for $y$ and $z$ using guess and check but we can also make this process a little easier as follows. If $z \geq 10$ then $2.17z > 20$ so this is not possible. From here, we can try the different possible values of $z$. For example, if $z = 7$, this gives $$0.61y + 2.17 \times 7 = 20$$ or $$y = \frac{20 - 2.17 \times 7}{0.61}\approx 7.88.$$ This is not a whole number and, since the number of pennies of weight $3.11$ grams must be a whole number, there can not be $7$ pennies of weight $4.67$ grams. Similar analysis will eliminate all other possible whole number values of $z$, between $0$ and $9$, except for $z = 5$ which gives us a value $$y = \frac{20 - 2.17 \times 5}{0.61}=15.$$ To check our answer, we observe that indeed $$0.61 \times 15 + 2.17 \times 5 = 20.$$ Thus there are $30$ pennies which weigh $2.50$ grams, $15$ pennies that weigh $3.11$ grams, and $5$ pennies that weigh $4.67$ grams.

2. In the case of $145$ grams, the only possibility which worked was $30$ pennies of weight $2.50$ grams, $15$ pennies of weight $3.11$ grams, and $5$ pennies of weight $4.67$ grams. Suppose more generally that two rolls of pennies have the same weight and that the numbers of pennies of each weight in these rolls are $x_1,x_2,x_3$ and $y_1,y_2,y_3$. Then we have $$x_1 + x_2 + x_3 = y_1 + y_2 + y_3 = 50$$ since each roll contains $50$ pennies. The two rolls have the same weight so this tells us that $2.50x_1 + 3.11x_2 + 4.67x_3 = 2.50y_1 + 3.11y_2 + 4.67y_3.$ Reasoning as in part (a) we find the equation $$0.61x_2 + 2.17x_3 = 0.61y_2 + 2.17y_3.$$ Rearranging this equality gives $$0.61(x_2-y_2) = 2.17(y_3-x_3).$$ Multiplying both sides by $100$ changes this into $$61(x_2-y_2) = 217(y_3-x_3).$$ The left hand side of this equation, $61(x_2 - y_2)$, is divisible by the prime number $61$. So the right hand side of the equation, $217(y_3 - x_3)$, must also be divisible by $61$. This is only possible if $y_3 - x_3$ is divisible by $61$. Since $y_3$ and $x_3$ are whole numbers between $0$ and $50$ we know that $$-50 \leq y_3 - x_3 \leq 50.$$ The only multiple of $61$ that meets these criteria is $0$. So it must be that $y_3 - x_3 = 0$ or $y_3 = x_3$. This then forces $x_2$ to be the same as $y_2$ and finally $x_1$ to be the same as $y_1$. So even in this situation with pennies of three different weights, the weight of the roll determines how many pennies of each type are in the roll.

3. We will continue to use $x_1,x_2,x_3$ for the number of pennies of respective weights $2.5,3.1,4.7$ grams in one roll and $y_1,y_2,y_3$ for the corresponding numbers in the second roll. The minimum weight for each roll is $125$ grams, achieved exactly when all of the pennies weigh $2.5$ grams. Beyond the $125$ grams each penny of weight $3.1$ grams contributes $0.6$ grams while each penny of weight $4.7$ grams contributes $2.2$ grams. So if the weights of these two rolls are the same this means $$0.6x_2 + 2.2x_3 = 0.6y_2 + 2.2y_3.$$ If we multiply both sides of this equation by $10$ to remove the decimals this gives $$6x_2 + 22x_3 = 6y_2 + 22y_3.$$ Dividing both sides by $2$ gives $$3x_2 + 11x_3 = 3y_2 + 11y_3.$$ One way to satisfy this equation is to choose $x_2 = y_2$ and $x_3 = y_3$ but this is the situation where there are the same number of pennies of weights $3.11$ and $4.67$ grams. So we are looking for a different solution. To solve the system in a different way we can simplify by taking $x_3 = y_2 = 0$, leaving an equation of the form $$3x_2 = 11y_3.$$ This equation has the whole number solution $x_2 = 11$ and $y_2 = 3$. Putting this all together we have a whole number solution to the equation with $x_2 = 11$, $x_3 = 0$, $y_2 = 0$, $y_3 = 3$. Going back and using the fact that $x_1 + x_2 + x_3 = y_1 + y_2 + y_3 = 50$ if $x_1 = 39$, $x_2 = 11$, and $x_3 = 0$ this has the same weight as $y_1 = 47$, $y_2 = 0$, $y_3 = 3$, the common weight being $131.6$ grams.