Coinbinations (1)
I want to buy a $1.00 newspaper. I have just enough pennies ($0.01), nickels ($0.05), dimes ($0.10), and quarters ($0.25) to buy the newspaper in every possible exact-change combination of those coins.
How many coins do I have?
Note: "Exact-change combination" doesn't necessarily mean that you must use more than 1 kind of coins listed above. For example, paying 4 quarters is allowed.
The answer is 134.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
14 solutions
But surely this answer is incorrect since the question specified a COMBINATION of coins so therefore there has to be two species of coins and so the most you need to ensure possible COMBINATIONS is 95 cent, 19 nickels, 9 dimes and 3 quarters - 126 coins. If it had said "exact-change possibility" then the 134 answer would be correct
Log in to reply
The question was ambiguous. One could also argue that one needed only 11 coins - 3 quarters, 1 Dime , 3 nickels and 5 pennies
Sorry I should have said AT LEAST two species of coin but this does nto affect the answer
at least one coin of each value gives 1 pennie + 1 nickel + 1 dime + 1 quarter = 41 cents
for reaching 100 , i need to sum 59 with either of these combinations
59 1 + 0 *5 + 0 10 + 0*25 = 59
4 1 + 11 5 + 0 10 + 0 25 = 59
9 1 + 0 5 + 5 10 + 0 25 = 59
9 1 + 0 5 + 0 10 + 2 25 = 59
wich gives
59 + 11 + 5 + 2 = 77 coins
+4 that i already have = 81 coins
Log in to reply
Yes, it's not clear. I understood that too, hat I needed to use every possible coin. So my first answer was 81 too.
The question could have been clearer
Log in to reply
I agree. In reality, trying to pay for a one dollar newspaper with 100 pennies is going to cause a lot of bad feeling; the transaction might be refused and the would-be buyer asked to leave the premises. And who's going to walk around with 134 coins in their pocket in loose change?
I agree with 81 because "combination of those coins". The problem is unclear
The correct answer is 81 coins...the operative instruction was a COMBINATION of the coins. Four quarters is not a combination of the coins.
Log in to reply
I agree that when you say a combination of coins most people would assume at least one of each coin, but I guess zero coins is mathematically correct for a combination as well. It still is misleading, so I got it correct in my second try.
I misunderstood for yet another reason. Now with my useless result, I am happy to claim that there is 242 exact-change combinations, and if you want to make each of these 242 exact-change combination, you will need 8864 coins.
There is a huge number of possibilities? Why are these not included? 1 sample comb: 0.1(5) + 0.25(2) + 0 + 0 = 1
Log in to reply
I misunderstood the same way you did, see my comment below. In fact, all of these combination was taken into account, but not separately. Hence, with the 100 pennies, 20 nickels, 10 dimes and 4 quarters, you can certainly do your example of combination.
But there is a note that she can also use only 1 type of coin
This was an intermediate question???
But yea you basically want 100 pennies 20 nickels, 10 dimes, and 4 quarters so you could pay with only one type of coin. Since you now have the maximum amount of coins you could also just use them to pay in any combination you wish.
100+20+10+4=134
I agree; I feel that intermediate was a bit too difficult. You can try this problem's sequel , though!
Log in to reply
Thank you Blan, Very cool!
Oh maybe my English is not good, I don't understand the problem totally. I thought the problem was "my coins can pay any exact number between 0.01 and 1.00" so I got the answer as 10 coins with 4 pennies, 1 nickel, 2 dimes, 3 quaters...... If I know what the problem means it certainly is easy. But I really can't understand with this statement
Log in to reply
Yes, unfortunately that interpretation is mistaken.
The stated goal was to actually purchase a physical item, costing $1.00
If you go to the store, it will do you no good to assemble any less than the actual price. For example, 40 cents would not buy a one-dollar item, so you don't need to worry about making 40 cents.
The part which probably confused you was when the problem said "in all possible combinations." What that means is that there are multiple ways to make a dollar (4 quarters, 5 dime plus 10 nickels, etc.); you need only enough coins to use every possible combination which can add up to one dollar.
Yeah not sure why it was intermediate
1 0 0 pennies are needed to make a dollar.
Any other combination of coins that makes up a dollar will use less than 1 0 0 pennies. This means we already have enough pennies for any combination.
2 0 nickels are needed to make a dollar on their own. Similarly, no combination of coins will use more than 2 0 nickels. Again, we have enough nickels.
Following the pattern, having 1 0 dimes and 4 quarters would fill their respective dollars, and no more would be needed for a different coin combination. Therefore 1 0 0 + 2 0 + 1 0 + 4 = 1 3 4 coins are sufficient.
Hmm, to me 'combination of these coins' implied there is at least one of each denomination in the payment!
The question was unclear. I interpreted the problem as having just enough of each of the coins to buy the $1.00 paper and be able to make every amount of change between 1 cent and 1 dollar. I came up with 5 pennies, 2 nickels, 1 dime and 3 quarters.
Log in to reply
That would make an interesting problem in itself.
I understand that ...
134 coins needed for 242 distinct combinations which would require 8864 coins to perform each distinct combination once
The problem with the problem is the question. The words “just” is misleading. A more clear presentation might be:
“I want to purchase a $1.00 paper. What is the maximum number of coins required to purchase the paper with any combination of pennies ($.01), nickels ($.05), dimes ($.10), and quarters ($.25)? Not all denominations have to be used in any combination.
Will in bath, ME
First, note that each different coin has a way to make 1.00 using only that specific coin.
Then the number of coins needed for that situation is 1.00/v, where v is the value of that coin. 1.00/v will be the maximum amount of unique coins needed because if any other coin is used then the remaining amount needed is less than 1.00, so the number of coins needed will be less than 1.00/v.
Since 1.00/v coins cover each coin's situation of only the specified coin, all possible situations will be covered using only the coins that cover the specified situation.
So the number of coins needed is 1.00/0.01+1.00/0.05+1.00/0.10+1.00/0.25 =100+20+10+4= 134 .
I like your generalization! To write it more formally, the amount of coins needed to pay $1 when there are c coins each with a different value of v n is n = 1 ∑ c ⌊ v n 1 0 0 ⌋ .
A general solution would be
⌊ 1 0 0 / 1 ⌋ + ⌊ 1 0 0 / 5 ⌋ + ⌊ 1 0 0 / 1 0 ⌋ + ⌊ 1 0 0 / 2 5 ⌋ where ⌊ ⌋ is the floor function
The question was how many coins to cover all possible combinations. 100 pennies is not a combination. A combination would include at least 1 coin of each denomination. Therefore consider 1 qtr;1 nickle ,1 dime =.40 + 60 pennies Largest number of pennies required. 2nd combo is 1 qtr; 1 nickle ; 6 dimes ; 10 pennies ( 6 dimes max number of dimes required ). 3rd 1 qtr ; 1 dime ; 5 pennies ; and 12 nickles max required 4th 3 qtrs the max ; 1 nickle ; 1 dime ; 10 pennies Recap 3 quarters; 6 dimes ; 12 nickles ; 60 pennies = 81 coins
my biggest problem is English, I solve the problems uses Google and this changes the excellence of the issue, I do not understand English, only Portuguese
You need 100 pennies, 20 dimes, 10 nickels and 4 quarters to make one dollar in each denomination.
Answer=(100+20+10+4)=134
You will need to cover the combinations: 1 0 0 pennies; 2 0 nickles; 1 0 dimes; 4 quarters. Therefore, you have 1 3 4 coins.
Seems trivial...why is this intermediate?
We will be assessing values in cents ($0.01).100 is divisible by all values of the coin (so each coin, by itself, can make a valid combination), and you can only place up to floor(100/x) of x valued coins (and floor(100/x)=100/x in this case as 100/x is an integer via the first statement) as any more would of said coin would cause there to be an excess value, so you can just sum the amount of coins needed to get 100 for each coin value x (100/x), for all x (which will be 134 when you do the calculations).
The answer can be seen when you isolate each coin and consider how many coins it takes to reach $1.00
100 pennines (worth $0.01 each) are needed for a dollar
20 nickels (worth $0.05 each) are needed for a dollar
10 dimes (worth $0.1 each) are needed for a dollar
4 quarters (worth $0.25 each) are needed for a dollar
Thus, the total coins you have is 100+20+30+4=134
They try to trick you here by getting you to consider all kinds of crazy combinations. Be sure to read the problem and know WHAT you're solving for before you get to the HOW, because you may be (as Stephen Covey puts it) exploring the wrong jungle.
I think the question can be misunderstood because it specifically stated “combination of those coins”. If each “combination” required one of each of the different coins at a minimum the correct answer would be 81. 60 pennies, 12 nickels, six dimes and 3 quarters. That’s also a slightly harder problem to solve.
This isn't a solution.
Hm. Not as hard as first thought. In fact almost trivial. I thought you had to use at least one of each of the four denominations of coin regardless of your combination. Now THAT is a bit more of an intermediate
100
pennies = $1.00.
20
nickels = $1.00.
10
dimes = $1.00.
4
quarters = $1.00.
You don’t have to think about the other combinations because you’ll never use more than that amount of pennies, nickels, dimes, or quarters to get $1.00.
Therefore, you would add 100 + 20 + 10 + 4 to get 1 3 4 .
Example:
100 pennies is already $1.00, if you use more in some combination, you’ll have more than $1.00. (Unless their cost is negative, but in this problem, we aren’t using negative integers).
I like your reasoning and example, but you didn't actually put the answer of 134 in your solution.
Log in to reply
Lol @Blan Morrison 😂😄
Notice that no change combination will require more than 100 pennies, because that would put us over one dollar. Similarly no change combination will require more than 20 nickels, more than 10 dimes or more than 4 quarters.
However to pay only with pennies requires 100 pennies, to pay only with nickels requires 20 nickels ,to pay only with dimes requires 10 dimes and to pay only with quarters requires 4 quarters.
Put these two observations together and you see that the number of coins required is
1 0 0 + 2 0 + 1 0 + 4 = 1 3 4