Question

How many ways are there to make 33 cents using any combination of pennies, nickels and dimes?

Answer 1

Hint: We have been given a problem related to the currency cents which has an even smaller denomination called penny, nickel and dime. In order to be able to represent cents in these smaller denominations, we must know how these different currencies are related to each other and also about how many pennies, nickels and dimes is one cent equivalent to. Then, we shall find the various combinations of these smaller denominations which add up to make 33 cents.

Complete step-by-step answer:
Given 33 cents along with penny, nickel and dime.
We know that penny, nickel and dime are smaller denominations of the currency cents.
From the worldwide acceptable currency system, we see that
1 penny = 1 cent
1 nickel = 5 cents
1 dime = 10 cents
Now, we shall list all the possible combinations of these smaller denominations which sum up to be equivalent to 33 cents.
Let P represent pennies, N represent nickels and D represent dimes.
Taking 3 dimes initially, we get
$3D+0N+3P=\left( 3\times 10 \right)+\left( 0\times 5 \right)+\left( 3\times 1 \right)$ cents
$\Rightarrow 3D+0N+3P=30+0+3$cents
$\Rightarrow 3D+0N+3P=33$cents
Taking 2 dimes, we get 3 combinations.
$2D+2N+3P=\left( 2\times 10 \right)+\left( 2\times 5 \right)+\left( 3\times 1 \right)$ cents
$\Rightarrow 2D+2N+3P=20+10+3$cents
$\Rightarrow 2D+2N+3P=33$cents
$2D+1N+8P=\left( 2\times 10 \right)+\left( 1\times 5 \right)+\left( 8\times 1 \right)$ cents
$\Rightarrow 2D+1N+8P=20+5+8$cents
$\Rightarrow 2D+1N+8P=33$cents
$2D+0N+13P=\left( 2\times 10 \right)+\left( 0\times 5 \right)+\left( 13\times 1 \right)$ cents
$\Rightarrow 2D+0N+13P=20+0+13$cents
$\Rightarrow 2D+0N+13P=33$cents
Taking 1 dime, we get 5 combinations.
$1D+3N+8P=\left( 1\times 10 \right)+\left( 3\times 5 \right)+\left( 8\times 1 \right)$ cents
$\Rightarrow 1D+3N+8P=10+15+8$cents
$\Rightarrow 1D+3N+8P=33$cents
$1D+2N+13P=\left( 1\times 10 \right)+\left( 2\times 5 \right)+\left( 13\times 1 \right)$ cents
$\Rightarrow 1D+2N+13P=10+10+13$cents
$\Rightarrow 1D+2N+13P=33$cents
$1D+1N+18P=\left( 1\times 10 \right)+\left( 1\times 5 \right)+\left( 18\times 1 \right)$ cents
$\Rightarrow 1D+1N+18P=10+5+18$cents
$\Rightarrow 1D+1N+18P=33$cents
$1D+0N+23P=\left( 1\times 10 \right)+\left( 0\times 5 \right)+\left( 23\times 1 \right)$ cents
$\Rightarrow 1D+0N+23P=10+0+23$cents
$\Rightarrow 1D+0N+23P=33$cents
Noticing the pattern, we observe that when we get 0 dimes, we get 7 combinations from taking 0 nickels to 6 nickels.
Therefore, the total number of combinations is $1+3+5+7=16$combinations.

Note: In many national currencies, the cent, commonly represented by the cent sign is a monetary unit that equals $\dfrac{1}{100}$ of the basic monetary unit. Etymologically, the word cent derives from the Latin word ‘centum’ meaning hundred. Cent also refers to a coin worth one cent.