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## Question

Guy walks into a coin shop and buys 100 coins and pays \$100 for them.

The purchase comprised a certain number of quarters and dollars at face value,

\$0.25 and \$1.00, respectively, and a certain number of gold coins known as Eagles.

What's the unit price of an Eagle?

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as far as i can see, there's not enough information.

firstly, the problem states nothing about the number of coins used to make the purchase, and even if it did, eagles could conceivably be any reasonable value.

we would further more need the ratio of quarters to dollars to eagles.

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As I recall, the man paid using a \$100 bill.

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It would seem from the information that the no of dollars (x), the number of quarters (y) and the number of eagles , face value z, quantity 100 -(x+ y) must be 100.

So 100 =1x + 0.25y = z(100 -(x+y)),

z =100-x-0.25y/100-x-y

I dont know if this can be simplified/ expanded further

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.

1. Eagles are worth an even number of whole dollars.
2. The number of Quarters equals the number of Dollars plus an even multiple of the number of Eagles.
.

With additional information comes a Bonus question:

What would be the value of an Eagle if it were an ODD number of whole dollars, the other information remaining unchanged?

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bonanova, here is the riddle stated the way i read it.

Guy walks into a coin shop and buys 100 coins [of unknown value] and pays \$100 for them.

The [payment of 100\$] comprised a certain number of quarters and dollars at face value,

\$0.25 and \$1.00, respectively, and a certain number of gold coins known as Eagles.

What's the unit price of an Eagle?

if he purchases the coins with a 100 dollar bill, then the eagle is worth 0.

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E + D + Q = 100

E*V(E) + D*V(D) + Q*V(Q) = 100

V(D) = 1

V(Q) = 0.25

V(E) = 2x + 1

Q = D + 2y*E

=> 2D + (2y+1)E = 100

since the variables are more than equation, hence heuristics can be used

=> at D=0, E=4 we have Q = 96

=> E*V(E) = 76

=> V(E) = 19

Edited by KlueMaster
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bonanova, here is the riddle stated the way i read it.

Guy walks into a coin shop and buys 100 coins [of unknown value] and pays \$100 for them.

The [payment of 100\$] comprised a certain number of quarters and dollars at face value,

\$0.25 and \$1.00, respectively, and a certain number of gold coins known as Eagles.

What's the unit price of an Eagle?

if he purchases the coins with a 100 dollar bill, then the eagle is worth 0.

Ah, I understand. OK. The OP states

"the purchase comprised a certain number of quarters and dollars ... and ... Eagles"

So the question of interpretation is what purchase [used here as a noun] refers to.

onelook.com gives us this:

Quick definitions from WordNet (purchase)

â–¸ noun: the acquisition of something for payment ("They closed the purchase with a handshake")

â–¸ noun: a means of exerting influence or gaining advantage ("He could get no purchase on the situation")

â–¸ noun: something acquired by purchase

It wouldn't be the first meaning, or the second. That leaves the third. That is, the Quarters, Dollars and Eagles are coins that were acquired by purchase. In none of the cases does purchase refer to something used as payment. In the puzzle, the payment was the \$100 bill.

So that's the intended sense of the puzzle - that certain numbers [non-zero numbers should be inferred] of Quarters, Dollars and Eagles, coins all, were purchased using a \$100 bill [which was exact payment.]

With the added information in my bad for leaving it out of OP, namely that the number of Quarters purchased equals the number of Dollars plus an even multiple of the number of Eagles and that the Eagle is worth a whole number, say n, of dollars, then there is a unique value for the Eagle coin [1] when n is odd and another unique value [2] when n is even.

Hope that clarifies.

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E + D + Q = 100

E*V(E) + D*V(D) + Q*V(Q) = 100

V(D) = 1

V(Q) = 0.25

V(E) = 2x + 1

Q = D + 2y*E

=> 2D + (2y+1)E = 100

since the variables are more than equation, hence heuristics can be used

=> at D=0, E=4 we have Q = 96

=> E*V(E) = 76

=> V(E) = 19

Although OP did not say this explicitly, the "certain numbers" of the three types of coins purchased are non-zero.

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X = number of dollars

Y = number of quarters

Z = number of eagles

A = face value of eagles

All variables in the following are non-negative integers.

Problem 1:

```
(1) X + Y + Z = 100

(2) X + Y / 4 + A*Z = 100

(3) A = 2 * B

(4) Y = X + 2 * C * Z

```
replacing Y from (4) in (1) and (2) and A from (3) in (2) we get:
```
(5) 2* X  + (2* C +1)* Z = 100

(6) X + X / 4 + C*Z/2 + 2*B*Z = 100

```
From (5) since X is an integer, Z must be even. Since Z is even and X, B and C are also integers, from (6) we get that X must be a multiple of 4.
```
(7) Z = 2 * D

(8) X = 4 * E

```
Replacing X and Z in equations (5) and (6) we get:
```
( 9) 8 * E + (2 * C + 1)* 2 * D = 100

(10) 5 * E + (C + 4 *B) * D = 100

```
From (9) we get that D is even since both 100 and 8*E are multiples of 4 and 2*C+1 is odd. From (10) because D is even, we get E is a multiple of 2
```
(11) D = 2 * F

(12) E = 2 * G

```
Rewriting (9) and (10) with this new information and dividing by 2 we get:
```
(13) 4 * G + (2 * C + 1)* F = 25

(14) 5 * G + (C + 4 *B) * F = 50

```
Clearly F is odd (from (13) since 25 is odd and 4*G is even). Rewriting previous equations (substracting 1 and dividing by 2 where possible) leaves us with:
```
(15) F = 2 * H + 1

(16) 2 * G + 2*C*H + C + H = 12

(17) 5 * G + 2*C*H + C + 8*B*H + 4*B = 50

```
From (16) we get C+H must be even. From (17) we get C+G must be even (since the rest of the right part is even and 50 is even). Replacing (16) in (17) we get:
```
(18) 3 * G + 12-H + 8*B*H + 4*B = 50

```
or equivalently:
```
(19) 3 * G + (8*B-1)*H + 4*B = 38

```
Clearly B=0 means A=0 hence Eagles would be for free, impossible. Therefore B>=1 so 2*B-1>=1>0 Same reasoning G>=1 (otherwise X=0). Hence (8*B-1)*H <= 38-3-4 = 31 Since B>=1, 8*B-1 >=7 so H <=floor(31/7)=4. Case 1) H=0 implies F=1; D=2; Z=4
```
(16') 2 * G + C = 12

(19') 3 * G + 4*B = 50

```
From (16'), since C+G is even, G must be even. Looking back at (16'), if G is even then C is a multiple of 4. From (19'), since 50 is not a multiple of 4, G cannot be a multiple of 4. Hence:
```
(20) C = 4 * I

(21) G = 4 * J+ 2

```
Rewriting (16) and (19) we get
```
(22) 2 * 4* J + 4 + 4*I = 12

(23) 3 * 4* J + 6 + 4*B = 50

```
Dividing by 4 and 2 respectively, and moving constants to the left part of the equations
```
(23) 2 * J + I = 2

(24) 3 * J + B = 11

```
From (22) we get J=0; I=2. From (24) we get B=11. Hence A=2*11=22. C=8; G=2; E=2*G=4; X=4*E=16; Z=2*D=4*F=4 Y=100-16-4 = 80 Verifying X+Y/4+16+80/4+22*4 is not 100. Case 2) H=1 implies F=2*H+1=3; D=2*F=6; Z=2*D=12
```
(16') 2 * G + 3*C = 11

(19') 3 * G + 12*B = 39

```
B>=1. Since G>=1 we get B<=(39-3)/12. So B<=3. - B=1 leads to G=(39-12)/3 = 27/3=9. However 2*9= 2*G <= 11 is a contradiction - B=2 leads to G=(39-24)/3 = 15/3=5. C=(11-2*5)/3=1/3 impossible - B=3 leads to G=(39-36)/3 = 3/3=1. C=(11-2)/3=3 Going backwards: X = 4*E = 8*G = 8 Y = 100-X-Y=100-8-12=80 = X+2*C*Z=8+6*12 A = 2*B=6 Verifying X + Y / 4 + A*Z = 8+ 80/4 + 6*12=100 So, the unit price for one gold Eagle is 6\$ Case 3) H=2 implies F=2*H+1=5; D=2*F=10; Z=2*D=20
```
(16') 2 * G + 5*C = 10

(19') 3 * G + 20*B = 40

```
B>=1. Since G>=1 we get B<=(40-3)/20. So B=1. Hence G=20/3 impossible. Case 4) H=3 implies F=2*H+1=7; D=2*F=14; Z=2*D=28
```
(16') 2 * G + 7*C = 9

(19') 3 * G + 28*B = 41

```
B>=1. Since G>=1 we get B<=(41-3)/28. So B=1. Hence G=(41-28)/3 = 13/3 impossible Case 5) H=4 implies F=2*H+1=9; D=2*F=18; Z=2*D=36
```
(16') 2 * G + 9*C = 8

(19') 3 * G + 36*B = 42

```

B>=1. Since G>=1 we get B<=(42-3)/36. So B=1.

Hence G=(42-36)/3 = 6/3=2.

C=(8-2*2)/9 impossible.

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X = number of dollars

Y = number of quarters

Z = number of eagles

A = face value of eagles

All variables in the following are non-negative integers.

Problem 1:

(1) X + Y + Z = 100
(2) X + Y / 4 + A*Z = 100
(3) A = 2 * B
(4) Y = X + 2 * C * Z

replacing Y from (4) in (1) and (2) and A from (3) in (2) we get:

(5) 2* X + (2* C +1)* Z = 100
(6) X + X / 4 + C*Z/2 + 2*B*Z = 100

From (5) since X is an integer, Z must be even.

Since Z is even and X, B and C are also integers, from (6) we get that X must be a multiple of 4.

(7) Z = 2 * D
(8) X = 4 * E

Replacing X and Z in equations (5) and (6) we get:

( 9) 8 * E + (2 * C + 1)* 2 * D = 100
(10) 5 * E + (C + 4 *B) * D = 100

From (9) we get that D is even since both 100 and 8*E are multiples of 4 and 2*C+1 is odd.

From (10) because D is even, we get E is a multiple of 2

(11) D = 2 * F
(12) E = 2 * G

Rewriting (9) and (10) with this new information and dividing by 2 we get:

(13) 4 * G + (2 * C + 1)* F = 25
(14) 5 * G + (C + 4 *B) * F = 50

Clearly F is odd (from (13) since 25 is odd and 4*G is even). Rewriting previous equations (substracting 1 and dividing by 2 where possible) leaves us with:

(15) F = 2 * H + 1
(16) 2 * G + 2*C*H + C + H = 12
(17) 5 * G + 2*C*H + C + 8*B*H + 4*B = 50

From (16) we get C+H must be even.

From (17) we get C+G must be even (since the rest of the right part is even and 50 is even).

Replacing (16) in (17) we get:

(18) 3 * G + 12-H + 8*B*H + 4*B = 50

or equivalently:

(19) 3 * G + (8*B-1)*H + 4*B = 38

Clearly B=0 means A=0 hence Eagles would be for free, impossible. Therefore B>=1 so 2*B-1>=1>0

Same reasoning G>=1 (otherwise X=0).

Hence (8*B-1)*H <= 38-3-4 = 31

Since B>=1, 8*B-1 >=7 so H <=floor(31/7)=4.

Case 1) H=0 implies F=1; D=2; Z=4

(16') 2 * G + C = 12
(19') 3 * G + 4*B = 50

From (16'), since C+G is even, G must be even.

Looking back at (16'), if G is even then C is a multiple of 4.

From (19'), since 50 is not a multiple of 4, G cannot be a multiple of 4.

Hence:

(20) C = 4 * I
(21) G = 4 * J+ 2

Rewriting (16) and (19) we get

(22) 2 * 4* J + 4 + 4*I = 12
(23) 3 * 4* J + 6 + 4*B = 50

Dividing by 4 and 2 respectively, and moving constants to the left part of the equations

(23) 2 * J + I = 2
(24) 3 * J + B = 11

From (22) we get J=0; I=2. From (24) we get B=11.

Hence A=2*11=22. C=8; G=2; E=2*G=4; X=4*E=16; Z=2*D=4*F=4 Y=100-16-4 = 80

Verifying X+Y/4+16+80/4+22*4 is not 100.

Case 2) H=1 implies F=2*H+1=3; D=2*F=6; Z=2*D=12

(16') 2 * G + 3*C = 11
(19') 3 * G + 12*B = 39

B>=1. Since G>=1 we get B<=(39-3)/12. So B<=3.

- B=1 leads to G=(39-12)/3 = 27/3=9. However 2*9= 2*G <= 11 is a contradiction

- B=2 leads to G=(39-24)/3 = 15/3=5. C=(11-2*5)/3=1/3 impossible

- B=3 leads to G=(39-36)/3 = 3/3=1. C=(11-2)/3=3

Going backwards:

X = 4*E = 8*G = 8

Y = 100-X-Y=100-8-12=80 = X+2*C*Z=8+6*12

A = 2*B=6

Verifying X + Y / 4 + A*Z = 8+ 80/4 + 6*12=100

So, the unit price for one gold Eagle is 6\$

Case 3) H=2 implies F=2*H+1=5; D=2*F=10; Z=2*D=20

(16') 2 * G + 5*C = 10
(19') 3 * G + 20*B = 40

B>=1. Since G>=1 we get B<=(40-3)/20. So B=1.

Hence G=20/3 impossible.

Case 4) H=3 implies F=2*H+1=7; D=2*F=14; Z=2*D=28

(16') 2 * G + 7*C = 9
(19') 3 * G + 28*B = 41

B>=1. Since G>=1 we get B<=(41-3)/28. So B=1.

Hence G=(41-28)/3 = 13/3 impossible

Case 5) H=4 implies F=2*H+1=9; D=2*F=18; Z=2*D=36

(16') 2 * G + 9*C = 8
(19') 3 * G + 36*B = 42

B>=1. Since G>=1 we get B<=(42-3)/36. So B=1.

Hence G=(42-36)/3 = 6/3=2.

C=(8-2*2)/9 impossible.

Bravo!

Care to find the unique odd-dollar value?

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.

1. Eagles are worth an even number of whole dollars.
2. The number of Quarters equals the number of Dollars plus an even multiple of the number of Eagles.
.

With additional information comes a Bonus question:

What would be the value of an Eagle if it were an ODD number of whole dollars, the other information remaining unchanged?

it seems to me there is more than one possible answer

Eagle worth \$4 (376 quarters, 2 dollars, 1 eagle) or (352 quarters, 4 dollars, 2 eagles)

Eagle worth \$8 (360 quarters, 2 dollars, 1 eagle) or (320 quarters, 4 dollars, 2 eagles).

I may be confused here though, so set me straight if I am answering the incorrect question

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Care to find the unique odd-dollar value?

Yes, of course

I was working on writing that, reusing previous notations and reasoning.

I wanted to mark the impact of the change from EVEN to ODD with red unfortunately that does not work inside code tags

X = number of dollars

Y = number of quarters

Z = number of eagles

A = face value of eagles

All variables in the following are non-negative integers

```
(1) X + Y + Z = 100

(2) X + Y / 4 + A*Z = 100

(3) A = 2 * B + 1

(4) Y = X + 2 * C * Z

```
replacing Y from (4) in (1) and (2) and A from (3) in (2) we get:
```
(5) 2* X  + (2* C +1)* Z = 100

(6) X + X / 4 + C*Z/2 + 2*B*Z + Z = 100

```
From (5) since X is an integer, Z must be even. Since Z is even and X, B and C are also integers, from (6) we get that X must be a multiple of 4.
```
(7) Z = 2 * D

(8) X = 4 * E

```
Replacing X and Z in equations (5) and (6) we get:
```
( 9) 8 * E + (2 * C + 1)* 2 * D = 100

(10) 5 * E + (C + 4 *B +2) * D = 100

```
From (9) we get that D is even since both 100 and 8*E are multiples of 4 and 2*C+1 is odd. From (10) because D is even, we get E is a multiple of 2
```
(11) D = 2 * F

(12) E = 2 * G

```
Rewriting (9) and (10) with this new information and dividing by 2 we get:
```
(13) 4 * G + (2 * C + 1)* F = 25

(14) 5 * G + (C + 4 *B+2) * F = 50

```
Clearly F is odd (from (13) since 25 is odd and 4*G is even). Rewriting previous equations (substracting 1 and dividing by 2 where possible) leaves us with:
```
(15) F = 2 * H + 1

(16) 2 * G + 2*C*H + C + H = 12

(17) 5 * G + 2*C*H + C + 8*B*H + 4*B + 4*H = 48

```
From (16) we get C+H must be even. From (17) we get C+G must be even (since the rest of the right part is even and 48 is even). Replacing (16) in (17) we get:
```
(18) 3 * G + 12-H + 8*B*H + 4*B + 4*H = 48

```
or equivalently:
```
(19) 3 * G + (8*B+3)*H + 4*B = 36

```
Clearly B=0 means A=0 hence Eagles would be for free, impossible. Therefore B>=1 Same reasoning G>=1 (otherwise X=0). Hence (8*B+3)*H <= 36-3-4 = 29 Since B>=1, 8*B+3 >=10 so H <=floor(29/10)=2. Case 1) H=0 implies F=1; D=2; Z=4
```
(16') 2 * G + C = 12

(19') 3 * G + 4*B = 36

```
From (19'), G must be even and B must be a multiple of 3. Looking back at (16'), if G is even then C is a multiple of 4. Hence:
```
(20) C = 4 * I

(21) G = 2 * J

(22) B = 3 * K

```
Rewriting (16) and (19) we get (dividing by 4 and 6 respectively)
```
(23) J + I = 3

(24) J + 2*K = 6

```
From (24) J is clearly even. Since J>0 (since X=16*J>0) and J<=3 this leads to J=2. Hence K=2; I=1; Moving backwards C=4; G=4; B=6; X = 16*J=32; Z=4 A = 2*B+1=13 Y = 100-32-4=64. Verifying X+Y/4+A*Z=32+64/4+13*4=32+16+52 = 100. So, the unit price for one gold Eagle is 13\$ Case 2) H=1 implies F=2*H+1=3; D=2*F=6; Z=2*D=12
```
(16') 2 * G + 3*C = 11

(19') 3 * G + 12*B = 33

```
B>=1. Since G>=1 we get B<=(33-3)/12. So B<=2. - B=1 leads to G=(33-12)/3 = 21/3=7. However 2*7= 2*G <= 11 is a contradiction - B=2 leads to G=(33-24)/3 = 9/3=3. C=(11-2*3)/3=5/3 impossible Case 3) H=2 implies F=2*H+1=5; D=2*F=10; Z=2*D=20
```
(16') 2 * G + 5*C = 10

(19') 3 * G + 20*B = 30

```

B>=1. Since G>=1 we get B<=(30-3)/20. So B=1.

Hence G=10/3 impossible.

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Unless I'm missing something, I get 2 solutions to the original problem:

If we define the following:

Q = # quarters

D = # dollars

E = # Eagles

V = Value of an Eagle

and write the solution as (Q,D,E,V) then my solutions are

(80, 8, 12, 6) which is araver's solution and

(80, 16, 4, 16)

So Eagle could be worth \$6 or \$16

For an odd value of V I get

(64, 32, 4, 13)

Eagle is worth \$13

edit: clarified target solution

Edited by HoustonHokie
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How's this: Eagles are \$15 a piece. Therefore, the dude bought: 3 Eagles for 45 dollars, 41 dollars (at \$41) and 56 quarters for \$14, totaling 100 coins for 100 dollars. How's that?

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How's this: Eagles are \$15 a piece. Therefore, the dude bought: 3 Eagles for 45 dollars, 41 dollars (at \$41) and 56 quarters for \$14, totaling 100 coins for 100 dollars. How's that?

The number of quarters (56) equals the number of dollars (41) plus an ODD multiple (5) of the number of Eagles (3). But the question said EVEN. Don't think that counts. If an ODD multiple is allowed, there are quite a few solutions possible...

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Unless I'm missing something, I get 2 solutions to the original problem:

If we define the following:

Q = # quarters

D = # dollars

E = # Eagles

V = Value of an Eagle

and write the solution as (Q,D,E,V) then my solutions are

(80, 8, 12, 6) which is araver's solution and

(80, 16, 4, 16)

So Eagle could be worth \$6 or \$16

For an odd value of V I get

(64, 32, 4, 13)

Eagle is worth \$13

edit: clarified target solution

You are correct:

I did not recheck case 1 which should read 38 instead of 50.

Hence there is a second solution for the first problem.

Case 1) H=0 implies F=1; D=2; Z=4

```
(16') 2 * G + C = 12

(19') 3 * G + 4*B = 38

```
From (16'), since C+G is even, G must be even. Looking back at (16'), if G is even then C is a multiple of 4. From (19'), since 38 is not a multiple of 4, G cannot be a multiple of 4. Hence:
```
(20) C = 4 * I

(21) G = 4 * J+ 2

```
Rewriting (16) and (19) we get
```
(22) 2 * 4* J + 4 + 4*I = 12

(23) 3 * 4* J + 6 + 4*B = 38

```
Dividing by 4 and 2 respectively, and moving constants to the left part of the equations
```
(23) 2 * J + I = 2

(24) 3 * J + B = 8

```

From (22) we get J=0; I=2. From (24) we get B=8.

Hence A=2*8=16. C=8; G=2; E=2*G=4; X=4*E=16; Z=2*D=4*F=4 Y=100-16-4 = 80

X+Y/4+A*Z=16+80/4+16*4 = 100.

So, the unit price for one gold Eagle is 16\$ in this case

Edited by araver
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Summary:

I wonder whether it's more fun to make these puzzles or to solve them...

First the close but no cigar awards ...

Preflop - yeah, but you forgot that Q + D + E = 100

KlueMaster - you forgot to read the OP's invisible ink section [my bad ] that said Q, D and E are non-zero.

Joey D - you get the most likely Google prowess award.

araver [repeating] the [claimed by OP to be unique] even answer.

araver the [actually] unique odd answer

The puzzle is adapted from a published one that asked the question in a simpler way.

I tried to toughen it up a bit. It gave the value of the Eagle [it wasn't about coins, but values

were given: \$0.25, \$1 and \$15] and asked how many of each would both total and cost 100.

That source might be known to Joey D, because he came up with the resulting 56 41 3 solution.

Which explains his award.

OK why did I claim the even solution was unique?

Because unconstrained there are a zillion solutions. My first try at limiting them was to specify

the value of the Eagle to be less than \$10. But that required further constraints. Then I spotted

the neat fact that [Q-D]/E = even integer eliminated all the cases except two - one made V odd,

the other made V even. Voila! I went with it.

Nice work all, and I exclude myself from that for the mistakes.

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