[Math] Pennies, Nickels, Dimes, and Quarters Summation of Money

algebra-precalculusproblem solving

Peter has only pennies, Norma only Nickels, Diane only dimes, and Quincy only quarters. Peter and Norma have the same number of coins, and Diane and Quincy have the same number of coins. What is the least number of coins they can all have if the sum of their money is \$4.87?

So far i have $p=n$, and $d=q$ as far as number of coins, but where do i go from here?

Best Answer

An extended description of the modular argument I gave above to reduce the case work. As per your request, consider the amount of money in terms of cents, not in terms of dollars.

Before properly defining what we mean by modular numbers, I will remind you of what we mean by an even number and an odd number. An even number is a whole number that you can divide by 2 and get another whole number back. (examples include 2,4,6,8,...). An odd number is one that if you divide by 2 you do not get a whole number back, but instead you get a remainder of one. (examples include 1,3,5,7,9,...).

Let me define a similar concept, which I will dub three-even, three-odd and three-other numbers. I call a number threeeven if when I divide by three, I get another whole number back (in tandem to how I define an even number. examples include 3,6,9,12,...). I call a number threeodd if when I divide by three, I get a remainder of one (like how I defined odd numbers. examples include 1,4,7,10,...). Finally, I define a new category, threeother where if I divide by 3 I get back a remainder of 2 (examples include 2,5,8,11,...).

This way of thinking of "even and odd", or "threeven, throdd, and throther", or even extending it further to "fourven, fodd, fother, fothing" or higher can be phrased mathematically as follows using modular notation:

A number $x$ is congruent to $y$ "modulo" $n$ (written as $x\equiv y\mod n$) when $x = y + kn$ for some integer $k$.

For example, $7 \equiv 1\mod 2$, since you have $7 = 1 + 3\cdot 2$. We can also say that $7\equiv 9\mod 2$ since you have $7=9+(-1)\cdot 2$. Often times, however, we prefer to have $0\leq y< n$ to make calculations easier.

Notice a few things:

  • if $x_1\equiv y_1\mod n$ and $x_2\equiv y_2\mod n$, then you have that $x_1+x_2\equiv y_1+y_2\mod n$
  • if $x_1\equiv y_1\mod n$ and $x_2\equiv y_2\mod n$, then you have that $x_1\cdot x_2\equiv y_1\cdot y_2\mod n$
  • if $x_1 = x_2$ then $x_1\equiv x_2\mod n$
  • $x+kn\equiv x\mod n$ for any $k$

The first property explains why an even number plus an odd number is always odd (since 0+1=1), just like how a throdd number plus a throdd number is always throther(since 1+1=2). The second property explains why an odd number times an odd number is always odd (since $1\cdot 1=1$), and why an even number times anything is even (since $0\cdot y = 0$)

Now with the definitions out of the way. Notice that $487\equiv 2\mod 5$ (since $487 = 485+2 = 5\cdot k + 2$ for some integer $k$ I am too lazy to compute). Notice also that $35 \equiv 0\mod 5$ (since $35 = 0+7\cdot 5$), and that $6\equiv 1\mod 5$.

Letting $p$ be the number of pennies (and therefore also the number of nickels), and $d$ be the number of dimes (and therefore also the number of quarters), you are given the equation:

$$6\cdot p + 35\cdot d = 487$$

We may now apply the third property from above and consider the equation above modulo 5.

$$6\cdot p + 35\cdot d \equiv 1\cdot p + 0\cdot d \equiv p \equiv 2\mod 5$$

Therefore, we know that $p$ will have a remainder of 2 when divided by 5. So, $p$ can only be one of the following numbers: 2,7,12,17,22,etc... (note: we chose mod5 because we got to completely cancel out the term where $d$ appears and replace it with zero. We could also have chosen to approach it via mod7, but division by 7 is not as quick and easy)

Continuing from the note above, and exploring the same with modulo 7: after some calculations you get: $6\equiv 6\mod 7$, $35\equiv 0\mod7$ and $487\equiv 4\mod 7$. (note, $6\equiv -1\mod 7$ from the fourth property above and is a bit easier to work with)

$$6p+35d\equiv 6p+0d\equiv 6p \equiv -1p\equiv 4\mod 7$$

so: $-1p\equiv 4\mod 7$ and $p\equiv -4\equiv 3\mod 7$, so $p$ is 3 more than a multiple of 7, $p$ has to be one of the numbers: 3,10,17,24,etc... Comparing with our list from above, $p$ has to be one of the numbers 17,52,87,122,etc...

We could figure out what the numbers have to be from there, but I will go through the process one more time from the other side. Consider each modulo 6. $6\equiv 0\mod 6$, $35\equiv 5\equiv -1\mod 6$, and $487\equiv 1\mod 6$.

$$6p+35d\equiv 0p+(-1)d\equiv -1d\equiv 1\mod 6$$

So, $d\equiv -1\equiv 5\mod 6$, and we know that $d$ has to be one of 5,11,17,23,etc...

Combining this information, we have then only a few cases to check in our range of numbers. 17 is too big for $d$, since already $17\cdot 35 = 595>487$. So, for $d=11$ we check and find that 17 works for $p$. So, noting finally that to use the fewest coins total implies we'll use the most possible dimes and quarters, the final answer is $p=n=17$ and $d=q=11$.

(final note, it does also show that you can get an amount of \$4.87 using 5 dimes and 5 quarters and 52 pennies and 52 nickels)

Related Question