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!
Interesting 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
BTW, regarding the new timer system, one ought to be able to see his own comments before the timer expires and the answers explode.
Answer = 159 cents = $1.59
9 pennies, 6 nickels, 2 dimes and 4 quarters.
9 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 = $1.59
Cool Puzzle man.
I think the answer is 9 pennys, 6 nickels, 2 dimes and 4 quarters.
Explanation:
Initially you have 5 variables and three equations:
1) P + 5N + 10D + 25Q = CHANGE
2) Q + 5P + 10N + 25D = CHANGE
3) P + N + D + Q = 21
USING (1) – (2) YOU GET:
4) -4P -5N -15D +24Q = 0
Now comes the tricky part for me:
You know that 4 is a factor of 24 and moreover that any number multiply by 4 or 24 would yield the same unit digit, so you have two options: P and Q are equal or P X a digit A gives you unit digit C and Q X digit B gives digit C too.
I explored option 1: Q and P are equal and it is impossible, because you get from (4):
5)-5N -15D +20Q = 0 from this you can get: N + 3D = 4Q OR 6) N/2 + 3D/2 = 2Q
And from (3) you get:
7) 2Q + D + N = 21
Now Substitute (6) in (7) you get 3N/2 + 5D/2 = 21 OR 3N + 5D = 42, Now you can noticed that N and D should be even so N+D would be even and equation 7 would be impossible.
From this P is not equal to Q
Option 2: 4 X digitA = 4X digitB, the only pairs of different digits that fulfill this are: (1 and 6),(2 and 7),(4 and 9)
Q less than P so going back to equation (4) you use this pairs and get:
With the first and second pairs is impossible to get 21 coins:
so the third pair works: -4×9 -5N -15D +24×4 = 0
5N + 15D = 60 The possible combination is N = 6 D= 2
and 9 + 6 + 2 + 4 = 21.
Cool!!.
Ohhh I forget the CHANGE would it be: 9 + 30 + 20 + 100= 159 cents: 1.59$
I 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.
p = # 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
$1.90
Nice 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.
$1.59
9 pennies
6 nickels
2 dimes
4 quarters
or
9 nickels
6 dimes
2 quarters
4 pennies
I 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?
@Hendy:
If you lay out the possibilities, you arrive to:
18x19x20/6=1,140 that is unless I have made a silly mistake