Is it possible to have 50 coins made up of pennies, dimes, and quarters that add up to $3?

Question

diaboloist99
You can substitute for one variable, but how would you go about solving for the other two?

Madhoo Palaka
This is not high school algebra. I'm looking for contradictions or proofs here. All the information is correct and in full.

Answer 1

If you try to make $3 with just quarters, you use 12 coins.
If you replace a quarter with dimes and pennies, you remove 1 coin but add 7 coins... a net change of 6 coins.

So no matter how you swap coins, you can't get to 50 coins. because 38 coins is not divisible by 6.

Now if you start with dimes you have 30 coins.
If you replace a dime with pennies, you remove 1 coin but add 10 coins... a net change of 9 coins.

Again, there is no way to get to 50 coins because 20 coins is not divisible by 9.

Finally if you try a combination of swaps:
Quarter to dimes and pennies --> +6 coins
Dimes to pennies --> +9 coins
Quarter to pennies --> +24 coins
Pennies to dimes --> -9 coins
Pennies to quarters --> -24 coins

As you can see the coin changes will always be multiples of 3, but you require 38 coins or 20 coins which are not multiples of 3.

Not possible.

Answer 2

The answer is no. I will give two solutions, the first provided by bill Cosby. Take 3 dollar bills to the bank and tell the teller you would like 50 coins in pennies, dimes, and quarters equal to your 3 dollars. He or she will not be able to come up with a combination that does the job. But of course we want to be sure the teller did not miss a combination of 50 coins that would equal 3 dollars. So let p, d, and q stand for the number (not the value) of pennies, dimes, and quarters respectively (each greater than 0) that add up to 50 and whose value is $3.00 if any such one or more combinations actually exist. Then we have two equations in three unknowns. The first equation expresses how many coins we must have: p+d+q=50, and the second equation expresses the value in dollars of the 50 coins: .01p+.10d+.25q=3.00. Multiplying the second equation through by 100 eliminates the decimals to give p+10d+25q=300. Subtracting the first equation from the last gives 9d+24q=250. Let's solve this equation for d by subtracting 24q and dividing by 9 to get d = (250-24q)/9. Now the number of quarters q is clearly 11 or less since 12 quarters is $3.00 eliminating the need for pennies or dimes which we wish to have among our change. So there is only a solution if q=1,2,3,4,5,6,7,8,9,10 or 11 quarters sovles the above equation for d and the solution is an integer, that is, a whole number of dimes. If you substitute in turn 1 through 11 for q and solve for d you will not get an integer solution for d. Therefore there is no integer solution for p+d+q = 50 and it is not possible to have 50 pennies, dimes, and quarters add up to $3.00. Note that this problem led to a system two equations in three variables so we could not solve it by the usual methods of elimination without a thrid equation.

Answer 3

OK - here is how to tell

Let x = pennies; y = dimes; z = quarter

You will get two equations.

x + y + z = 50
.01x + .1y + .25z = 3.00

Now, it gets a little complicated since you have no additional information. So algebraically, I think this becomes very difficult. However, we can say the following: In order to get 0 at the end of the figure (3.00) you must have an odd number of pennies (5) and odd number of quarters (5) - or an even number of quarters (0). If you have an even number of quarters, you would then have to have both dimes and pennies be odd or both be even. So let O by odd and E be even:

There are 8 combinations of coins (O, E)

1. OOO
2. OOE
3. OEO
4. OEE
5. EOO
6. EOE
7. EEO
8. EEE

In order to have an even result of coins, combo 1, 4, 6, 7 are out. That leaves combos 2, 3, 5, 8. However, in order to get 3.00, you must have both pennies and quarters be even or odd, so combo 2 and 5 are out. That leaves you with the following combos 3, 8:

O + E + O = 3.00
E + E + E = 3.00

If you say you must have at least one of each coin, then the minimum number of pennies is 5. Also, you cannot have more than 10 quarters; since 11 quarters would be 2.75 and you would not be able have 50 coins.

So quarter 11
5 pennies + x.10 +1q =3.00; 6 coins, leave too many dimes. This combo is out.
15 pennies + x.10 + 1q =3.00 16 coins, leave too many dimes. This combo is out.
25 pennies + x.10 + 1q = 3.00 26 coins, leave too few dimes. This and all other combos for 1 quarter is out.

Let’s check 2 quarters:

So quarters 11
10pennies + x.10+ 2q = 3.00; 12 coins, leaves too many dimes. This combo is out.
20pennies + x.10 +2q = 3.00; 22 coins, leaves too many dimes. This combo is out.
30 pennies + x.10 +2q = 3.00; 32 coins, leave too few dimes. This and all other combos for 2 quarter is out.

Let’s try 3

So now we have 11
5 pennies + x.10 +3q =3.00; 8 coins, leave too many dimes. This combo is out.
15 pennies + x.10 + 3q =3.00; 18 coins, leave too many dimes. This combo is out.
25 pennies + x.10 + 3q = 3.00; 28 coins leave too many dimes and all other combos for 3 quarters will leave too few dimes. .

Let’s do 4 quarters:
So quarters 11
10pennies + x.10+ 4q = 3.00; 14 coins, leaves too many dimes. This combo is out.
20pennies + x.10 +2q = 3.00; 24 coins, leaves too many dimes. This combo is out.
30 pennies + x.10 +2q = 3.00; 34 coins, leave too few dimes. This and all other combos for 2 quarter is out.

So quarters = 11 5 pennies + x.10 +5q =3.00; 10 coins, leave too many dimes. This combo is out.
15 pennies + x.10 + 5q =3.00 20 coins, leave too many dimes. This and all other combos for 5 quarters will leave too many dimes.

If you do this same process for 6,8,10 and 5,7,9 you will end up with the same result.

Answer – it cannot be done if you require at least one of each coin.

Answer 4

Pennies And Quarters

Answer 5

p=number of pennies, non-neg integer
d=number of dimes, non-neg integer
q=number of quarters, non-neg integer

p+d+q=50
p+10*d+25*q=300

There are some hidden constraints/limits here, such as:

p/5 = integer (the number of pennies must be evenly divisible by 5)

d<30 (because 30 dimes is $3)

q<12 (because 12 quarters is $3)
q can be further reduced to <11

q+d>10 (because 40 pennies is too small)

p/5 + q = even number (increments of 5 cents must balance)

I think this gives enough information to solve, but it's been way too long for me :) I can tell you, though, that 51 coins can equal $3: 40 pennies, 1 dime, 10 quarters

Answer 6

p = number of pennies
d = number of dimes
q = number of quarters
so
p + d + q = 50
p + 10d + 25q =300

Use linear algebra and elementary row operations to put this in row reduce form:

p - (5/3)q = (200/9)
d + (8/3)q = (250/9)

Which also means
p = (200/9) + (5/3)q
d = (250/9) - (8/3)q

We want whole numbers, so multiply both sides of both equations by 9:
9p = 200 + 15q
9d = 250 - 24q

There are an infinite number of solutions for this system...but we can only have whole numbers when it comes to numbers of coins. There are no solutions to this system. Why? Because 200 + 15q would have to be divisible by 9 for where q is an integer, and this is impossible. Same for 250 - 24q.

Answer 7

x -> number of pennies
y -> number of dimes
z -> number of quarters
then
x+ y+ z = 50
x + 10y + 25z =300
but you probably forgot some other piece of information, because you have three variables and only two equations. You need three equations!

Answer 8

I fed it into excel w/ the parameters as stated... does not work w/ whole coins... closest you can get is $2.90 using 40 pennies and 10 quarters.

Answer 9

x+y+z=50
x+10y+25z=300
Solve for x, y and z

Answer 10

yes it is possible

0.25q + 0.10d + 0.01p = $3.00
q + d + p = 50

work it out