## Hard math puzzle, LOOSE CHANGE

Dylan looked at his loose change and saw four types of coins: pennies, nickels, dimes, and quarters. After studying the coins, he made the following comment:

Wow; if my pennies were nickels, my nickels were dimes, my dimes were quarters, and my quarters were pennies, I’d still have the same amount of money as I do now.

If the Dylan had 21 coins, how much money was his change worth?

Thanks to Bobo The Bear for submitting this original math puzzle!

*Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!*

Hex| PUZZLE MASTER | Profile September 8th, 2010 - 3:30 amInteresting puzzle

P + N + D + Q = 21 (1)

1 <= P <= 18

1 <= N <= 18

1 <= D <= 18

1 <= Q 4P + 5N + 15D – 24Q = 0 (2)

(1) & (2) -> N = 28Q – 11D – 84 ->

28Q – 11D – 84 >= 1 -> D 1 Q >= 96/28 -> Q >= 4

28Q – 11D – 84 D >= (28Q – 102)/11 -> 18 >= (28Q – 102)/11 -> Q Q <= 10

So 4 <= Q D = (29Q + P -105)/10

For each value of Q between 4 and 10, we test what values between 1 and 18 for P give an integer D. We further check that (1) is satisfied:

P N D Q Value(cents)

9 6 2 4 159

10 1 5 5 190

1 7 7 6 256

2 2 10 7 287

So 4 possible solutions

Hex| PUZZLE MASTER | Profile September 8th, 2010 - 3:40 amBTW, regarding the new timer system, one ought to be able to see his own comments before the timer expires and the answers explode.

munna| Profile September 8th, 2010 - 7:30 amAnswer = 159 cents = $1.59

9 pennies, 6 nickels, 2 dimes and 4 quarters.

Shawn| PUZZLE GRANDMASTER | Profile September 8th, 2010 - 8:23 am9 pennies

6 nickels

2 dimes

4 quarters

9*.01 + 6*.05 + 2*.10 + 4*.25 = .09 + .30 + .20 + 1.00 = $1.59

9*.05 + 6*.10 + 2*.25 + 4*.01 = .45 + .60 + .50 + .04 = $1.59

suineg| PUZZLE MASTER | Profile September 8th, 2010 - 8:26 amCool Puzzle man.

I think the answer is 9 pennys, 6 nickels, 2 dimes and 4 quarters.

Explanation:

Initially you have 5 variables and three equations:

1) P + 5N + 10D + 25Q = CHANGE

2) Q + 5P + 10N + 25D = CHANGE

3) P + N + D + Q = 21

USING (1) – (2) YOU GET:

4) -4P -5N -15D +24Q = 0

Now comes the tricky part for me:

You know that 4 is a factor of 24 and moreover that any number multiply by 4 or 24 would yield the same unit digit, so you have two options: P and Q are equal or P X a digit A gives you unit digit C and Q X digit B gives digit C too.

I explored option 1: Q and P are equal and it is impossible, because you get from (4):

5)-5N -15D +20Q = 0 from this you can get: N + 3D = 4Q OR 6) N/2 + 3D/2 = 2Q

And from (3) you get:

7) 2Q + D + N = 21

Now Substitute (6) in (7) you get 3N/2 + 5D/2 = 21 OR 3N + 5D = 42, Now you can noticed that N and D should be even so N+D would be even and equation 7 would be impossible.

From this P is not equal to Q

Option 2: 4 X digitA = 4X digitB, the only pairs of different digits that fulfill this are: (1 and 6),(2 and 7),(4 and 9)

Q less than P so going back to equation (4) you use this pairs and get:

With the first and second pairs is impossible to get 21 coins:

so the third pair works: -4×9 -5N -15D +24×4 = 0

5N + 15D = 60 The possible combination is N = 6 D= 2

and 9 + 6 + 2 + 4 = 21.

Cool!!.

suineg| PUZZLE MASTER | Profile September 8th, 2010 - 10:23 amOhhh I forget the CHANGE would it be: 9 + 30 + 20 + 100= 159 cents: 1.59$

Jimmy Anders| PUZZLE MASTER | Profile September 8th, 2010 - 2:43 pmI get 3 answers.

Let:

p be the number of pennies,

n be the number of nickels,

d be the number of dimes, and

q be the number of quarters.

Then Dylan’s coins are worth p + 5n + 10d + 25q in cents, and p + n + d + q = 21.

The second paragraph (in the problem) means that:

p + 5n + 10d + 25q = 5p + 10n + 25d + q. Combining the terms:

4p + 5n + 15d = 24q. *

Then, notice that the coefficients on the left add up to 24. This means that whatever we choose for q, that p = n = d = q will solve eqn *, and from that choice you just have to “make change” to satisfy p + n + d + q = 21.

Then to make change, notice that reducing d by 1 and increasing n by 3 maintains equality in *, but increases the coin count by 2, and reducing n by 4 and increasing p by 5 maintains equality while increasing the coin count by 1 (naturally going to opposite way has the opposite effect on the count).

This way I found answers for when q is 4, 5, or 6. These are

p = 9, n = 6, d = 2, q = 4 => $ = $1.59,

p = 10, n = 1, d = 5, q = 5 => $ = $1.90,

p = 1, n = 7, d = 7, q = 6 => $ = $2.56.

MarkS| Profile September 9th, 2010 - 4:02 pmp = # of pennies, n= # of nickels, d = # dimes, q = # of quarters so we have:

p+n+d+q = 21 or q = 21-p-n-d and

p+5n+10d+25q = 5p+10n+25d+q or 4p+5n+15d-24q = 0, substituting for q gives

4p+5n+15d-24(21-p-n-d) = 0 or 28p+29n+39d = 504. then I got lucky and noticed

with p=n=2 the equation becomes 39d=390 which has a nice solution and we end up with

p=2, n=2, d=10 and q = 7 so he has $2.87

brianu| Profile September 9th, 2010 - 11:00 pm$1.90

Bobo The Bear| PUZZLE MASTER | Profile September 10th, 2010 - 12:03 pmNice job to all.

I had found the $1.59 solution, and I guess there was a glitch in my method, because I had honestly thought that the solution was unique. Congrats to all who found multiple solutions. The next time, I’ll have to be more exhaustive. Lesson learned.

Hendy| Profile September 15th, 2010 - 10:34 am$1.59

9 pennies

6 nickels

2 dimes

4 quarters

or

9 nickels

6 dimes

2 quarters

4 pennies

Hendy| Profile September 17th, 2010 - 8:56 amI went to the trouble to fill in a spreadsheet with all the possible combinations I could come up with for coins (1,114 by my clumsy method). I was able to show HEX’s answers were the only ones I found:

9,6,2,4 = $1.59

10,1,5,5 = $1.90

1,7,7,6 = $2.56

2,2,10,7 = $2.87

Does anyone have the equation for the possible combinations?

Hex| PUZZLE MASTER | Profile September 17th, 2010 - 11:48 am@Hendy:

If you lay out the possibilities, you arrive to:

18x19x20/6=1,140 that is unless I have made a silly mistake