Math Olympiad Problem Involving Currency

Question

A man has 2 pennies, 3 nickels, 1 dime, and 2 quarters. How many different sums of money can he make using one or more of these 8 coins?

I know the answer It's simple, however, the Olympiad requires you to answer it in under 1 min which is very hard considering you're given 1 minute or less to answer questions like these. Please suggest a quick method.

Answer 1

Adding (or not, if 0) coins in order from greater to lower value. In the diagram below:

In the first line of totals included 0 or 1 or 2 quarters.

In the second line of totals included 0 or 1 dimes.

In the third line of totals included 0 or 1 or 2 or 3 nickels.

Then, in the third line, eliminate (strike through) duplicate sums. (8 eliminated).

0 25 50

/ \ / \ / \

0 10 25 35 50 60

/ / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \

0 5 10 15 10 15 20 25 25 30 35 40 35 40 45 50 50 55 60 65 60 65 70 75

We got 16 different sums so far.

Now, adding 0 or 1 or 2 pennies, we notice that any number of added pennies produces different sums with any of the previously reached different 16 sums above.

So there will be 16 * 3 = 48 different resulting sums.

As the question is about using one or more coins, the leftmost branch of the diagram, with adding 0 pennies, means there are 0 coins. This case should be excluded 48 - 1 = 47.

Answer: Using one or more of these 8 coins, he can make 47 different sums of money.

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If all the reasoning was done mentally, one would need (I certainly would) to put only numbers on scrap paper. So before writing the answer, 47, or selecting it, if that's a multiple choice, my work would look like the three lines of numbers below, that is, even without 16 * 3 = 48 => - 1 = 47. If that was a new type of a problem for me on the test, I would not be able to do this in 1 minute or 3 minutes, or ..., but that's possibly the fastest way to solve it.

0 25 50

0 10 25 35 50 60

0 5 10 15 10 15 20 25 25 30 35 40 35 40 45 50 50 55 60 65 60 65 70 75

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For the last note, I first tried a couple of more approaches for the speediest one. One of them seemed promising until halfway.

From 256 cases, eliminate duplicates caused by exactly 1 of the 2 pennies in the sum, because having one or another of them would end up with the same sum, everything else being equal. Those are 64.

Similarly, with 2 quarters. 64.

Similarly, with 3 nickels. 128.

256 - 64 - 64 - 128 = 0.

Then, it should be understood, when we were eliminating duplicate-with-pennies cases, some of the duplicate-with-quarters cases were eliminated too. So after eliminating the latter, we should add back the twice-removed.

Similarly deal with eliminating cases twice while eliminating pennies-cases and nickels-cases independently.

Similarly deal with eliminating cases twice while eliminating quarters-cases and nickels-cases independently.

I got with numbers that far relatively quickly, but then the forward-reasoning led me to anticipating more steps, at least two:

1. Eliminating cases which were returned more than once (!) while returning the previously twice-removed ones.

2. Eliminating cases where the dime can be substituted by two of the three nickels and where a quarter can be substituted by the dime and three nickes, while taking care of the double-elimination(!) along with prior eliminations.

Once I got to anticipating the step 2 above, about substitution (!), I got the idea of the "featured" :) solution.

Answer 2

45 or 46

p-n-d-q

2-3-1-2= 77 cents

1-000= 1

2-000=2

0100=5

0010=10

0200=10

1010=11

2010=12

0110=15

16

17

20

21

22

25

26

27

30

31

32

35

36

37

40

41

42

45

46

47

50

51

52

55

56

57

60

61

62

65

66

67

70

71

72

75

76

77

6x7+3=45 different amounts

just count them

6 70's

6 60's

6 50's

6 40's

6 30's

6 20's

6 10's

3 units

46 if you also count 0 of each

Answer 3

A man has 2 pennies, 3 nickels, 1 dime, and 2 quarters. How many different sums of money can he make using one or more of these 8 coins?

Review the word problem and sort out the coins.

The possible number of each coin are:

Pennies: 0,1,2 (3 options)

Nickels: 0,1,2,3 (4 options)

Dimes: 0,1 (2 options)

Quarters: 0,1,2 (3 options)

Calculate the total number of combinations and subtract the case where no coins are selected.

Step 1

Multiply the number of options for each coin:

3 x 4 x 2 x3 = 72

Step 2

Subtract 1 from the total (the case where no coin is selected)

72 - 1 = 71

Solution:

There are 71 different sums of money that you can be combinate. However, after deducting multiple duplicates the answer is 47 different sums of money that you can be combinate.

Proof of answer:

After eliminating all the duplicates the final total is 47

1, 2, 5, 6, 7, 10, 11, 12, 15, 16, 17, 20, 21, 22, 25, 26, 27, 30, 31, 32, 35, 36, 37, 40, 41, 42, 45, 46, 47,

50, 51, 52, 55, 56, 57, 60, 61, 62, 65, 66, 67, 70, 71, 72, 75, 76 and 77.

I hope the mathematical calculations (step by step approach) was helpful. Please let me know if you require any more assistance. If anyone in my neighborhood is interested in setting up an in-person math tutoring session. I look forward to hearing from them. Have an amazing day. Doris H.

Answer 4

Hey L.C.! I saw Dr. Turner's answer, and she explained the structured approach really well. If you're looking for a way to apply it even faster in a competition setting, I totally get that pressure—sometimes you just don’t have time to list out everything systematically.

One trick that helps is recognizing how the coins build on each other. Instead of manually writing every sum, start by thinking in layers. You’ve got pennies (1, 2), then nickels (5, 10, 15), then the dime (which overlaps with 10 but also pushes things up to 20, 25, etc.), and finally quarters (25, 50). If you keep a running mental list and focus on what new numbers you can reach at each step, you can avoid double-counting while quickly estimating the range of possible sums.

Another speed hack is recognizing that since there are 8 coins, the total number of subsets is 255 (excluding the empty set), but because of overlap, the number of unique sums is much smaller—31, as Dr. Turner pointed out. Instead of thinking about combinations one by one, you can get a sense of the spread and quickly narrow it down to a ballpark answer.

If you want to get really fast at this, try practicing with different sets of coins and see how quickly you can mentally generate the possible sums. Over time, it becomes more intuitive, and you won’t even need to think about individual cases—you’ll just "see" the answer. Let me know if you want to run through another example together!

Answer 5

The smallest possible sum is 1 (one penny).

The largest possible sum is 77 (all coins used).

The trick is to eliminate all impossible sums between 1 and 77. The fact that there are only two pennies available allows for this elimination.

For example, you can not have a sum of 3 or a sum of 4 (not enough pennies).

You also cannot have a sum of 8 or 9 (1 nickel, but not enough pennies).

So from 1-9 eliminate 3,4,8,9

From 10-19, eliminate 13,14,18,19

From 20-29, eliminate 23,24,28,29

From 30-39, 4 sums eliminated

From 40-49, 4 eliminated

From 50-59, 4 gone

From 60-69, another 4 eliminated

From 70-77, eliminate 73, 74

That's 7(4) + 2 or 30 eliminations from 77 possibilities.

That means there are 77 - 30 or 47 possible sums that can be formed from the 8 coins.

Answer 6

A fast way to solve this problem is to use the "constructive counting" method, which involves you listing sums systematically while avoiding repetition.

Step-by-Step Quick Method

1. Identify Coin Values
2. Pennies (P): 1, 1
3. Nickels (N): 5, 5, 5
4. Dime (D): 10
5. Quarters (Q): 25, 25
6. Use a Binary Representation Approach : Each coin can either be included or not in a sum. With 8 coins, there are 28−1=2552^8 - 1 = 25528−1=255 possible subsets (excluding the empty set), but many will give the same total.
7. Generate Sums Systematically
8. Start from 0 and add one coin at a time, ensuring unique sums:
9. Pennies: 1, 2
10. Nickels: 5, 10, 15
11. Dime: 10
12. Quarters: 25, 50
13. Construct Unique Totals
14. Begin with the lowest-value coins and systematically add them while ensuring no duplicates:
15. Possible sums:
16. {1, 2, 5, 6, 7, 10, 11, 12, 15, 16, 17, 20, 21, 25, 26, 27, 30, 31, 32, 35, 36, 40, 41, 45, 46, 50, 51, 55, 56, 60, 66}
17. Final Answer: 31 different sums

How to Do It in Under 1 Minute?

1. Recognize it’s a subset sum problem.
2. Use small numbers first, then build combinations quickly.
3. Keep a systematic list, ideally using an organized mental pattern.
4. Avoid repetition—each sum should be counted once.

This method helps you avoid missing sums and prevent double-counting while ensuring speed. I hope this helps. If you have any additional questions, please let me know.

Take care!

Dr. Christal-Joy Turner