There isn’t a formula for this kind of problem; you’ll have to count up the options. But this could take forever! Pay attention to patterns to help the math go faster.
One of each type must be included, so that’s 1 quarter, 1 dime, 1 nickel, and 1 penny, for a total of 41 cents. Notice anything?
The desired final sum is 110 cents. If you have one penny, then you must have at least 5 pennies; otherwise, it’s impossible to get to a units digit of 0 in the final sum.
So you have to have a minimum of 1 quarter, 1 dime, 1 nickel, and 5 pennies, for a total of 45 cents. There are still 65 cents to go. The real question is how many ways you can make up this 65 cents.
Start with the largest denomination (quarters) and try to maximize. If you have two more quarters, or 50 cents, then how many ways are there to get the remaining 15 cents? You could have:
- 1 dime and 1 nickel
- 1 dime and 5 pennies
- 3 nickels
- 2 nickels and 5 pennies
- 1 nickel and 10 pennies
- 15 pennies
That’s a total of 6 possibilities if you start with 2 quarters.
What if you have just 1 quarter? Then you’d need another 40 cents to get to 65 cents. How many ways are there to get 40 cents, using just one quarter?
- 4 dimes
- 3 dimes and another 10 cents: either 2 nickels, 1 nickel and 5 pennies, or 10 pennies (3 options total)
- 2 dimes and another 20 cents: either 4 nickels, 3 nickels…
Noticing anything here? It’s actually not necessary to figure out the number of pennies each time. Just note that there’s only one possibility for each of the numbers of nickels (plus the corresponding number of pennies): 4 nickels, 3 nickels, 2 nickels, 1 nickel, and 0 nickels. That last possibility is the same as saying “all pennies.”
In other words, there are 5 ways to have 2 dimes and another 20 cents.
Back to the list:
- 1 dime and another 30 cents: either 6, 5, 4, 3, 2, 1, or 0 nickels (7 options total)
- 8, 7, 6, 5, 4, 3, 2, 1, or 0 nickels (9 options total)
So there are 1 + 3 + 5 +7 + 9 = 25 ways to get 65 cents using just one quarter. (How can you add that quickly? Remember the rule for summing consecutive numbers: the average times the number of terms equals the sum. In this case (5)(5) = 25.) Also, that’s interesting…the number of options is consecutive odds. Hmm.
So far, there are 6 + 25 = 31 ways to get 65 cents (using either 2 quarters or just 1).
What if you don’t use any quarters at all? You could have:
- 6 dimes and another 5 cents (2 ways to do this)
- 5 dimes and another 15 cents: either 3 nickels, 2 nickels, 1 nickel, or 0 nickels (4 options total)
- 4 dimes and another 25 cents: either 5, 4, 3, 2, 1, or 0 nickels (6 options total)
Noticing the pattern? It keeps going up by 2 each time, just like last time!
- 3 dimes = 8 options total
- 2 dimes = 10 options total
- 1 dime = 12 options total
- 0 dimes (that is, all nickels and pennies: 13 nickels, 12 nickels, ... 0 nickels) = 14 options total
Note that the very last option of the last row, 0 nickels, is the equivalent of having 0 quarters, 0 dimes, 0 nickels, and 65 pennies. In other words, you’ve covered all possible options.
So there are 2 + 4 + 6 + 8 + 10 + 12 + 14 = (8)(7) = 56 ways to get to 65 cents not using any quarters at all.
Therefore, there are 31 + 56 = 87 total possible ways to get the desired 65 cents.
The correct answer is (D).
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we shall fight on the beaches,
we shall fight on the landing grounds,
we shall fight in the fields and in the streets,
we shall fight in the hills;
we shall never surrender!