SOLUTION: How many ways can you make \$0.25 using pennies, nickels, dimes, and quarters

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 Question 257726: How many ways can you make \$0.25 using pennies, nickels, dimes, and quarters Found 3 solutions by drk, richwmiller, Edwin McCravy:Answer by drk(1908)   (Show Source): You can put this solution on YOUR website!There appear to be 13 ways. Here they are in {p,n,d,q} form: {p,n,d,q} {0,0,0,1} (0,1,2,0} {5,0,2,0} {0,3,1,0} {5,2,1,0} {10,1,1,0} {15,0,1,0} {0,5,0,0} {5,4,0,0} {10,3,0,0} {15,2,0,0} {20,1,0,0} {25,0,0,0} Answer by richwmiller(10275)   (Show Source): You can put this solution on YOUR website!one way with q 1q 2 dimes and a nickel 2 dimes and 5 pennies 1 dime 3 n 1d 2n 5p 25= 1q=1 +(2d+5) 1n or 5p 2 +(1d+15) n+10p 2n+5p 2 + 5n 1 +(4n+5) 1 +(2n+15) d+5p 1 +25p 1 +20p+5 1 +15p+10 +d +2n 2 +10p+15 d+n 1 +(5p+20) 2d 1d+2n 2 15 if various combos of individual pennies, nickels and dimes are counted as 1 Not interested in counting if each penny is counted as a separate way. Then the 25 pennies become 25! So there is probably some duplication here . Answer by Edwin McCravy(9716)   (Show Source): You can put this solution on YOUR website!``` p + 5n + 10d + 25q = 25 There is just one way to have 25 cents with only 1 quarter. 1. 1 quarter = \$0.25 So for all the other ways we only have pennies, nickels and dimes. p + 5n + 10d = 25 The number of pennies must be a multiple of 5 (including 0) So let p = 5k where k is an integer 5k + 5n + 10d = 25 Divide through by 5 k + n + 2d = 5 k + n = 5 - 2d So d can only be 0 thru 2 and have the right side positive. For d = 0, we have k + n = 5, there are 6 possibilities, k = 0 thru 5 and n = 5 thru 0 2. k=0, 0 pennies, 5 nickels, 0 dimes = \$0.25 3. k=1, 5 pennies, 4 nickels, 0 dimes = \$0.25 4. k=2, 10 pennies, 3 nickels, 0 dimes = \$0.25 5. k=3, 15 pennies, 2 nickels, 0 dimes = \$0.25 6. k=4, 20 pennies, 1 nickels, 0 dimes = \$0.25 7. k=5, 25 pennies, 0 nickels, 0 dimes = \$0.25 For d = 1, we have k + n = 3, there are 4 possibilities, k = 0 thru 3 and n = 3 thru 0 8. k=0, 0 pennies, 3 nickels, 1 dime = \$0.25 9. k=1, 5 pennies, 2 nickels, 1 dime = \$0.25 10. k=2, 10 pennies, 1 nickels, 1 dime = \$0.25 11. k=3, 15 pennies, 0 nickels, 1 dime = \$0.25 For d = 2, we have k + n = 1, there are 2 possibilities, k = 0 thru 1 and n = 1 thru 0 12. k=0, 0 pennies, 1 nickel, 2 dimes = \$0.25 13. k=1, 5 pennies, 0 nickels, 2 dimes = \$0.25 So the total is 1 + 6 + 4 + 2 = 13 ways. Edwin```