What is the exact max amount of coins each (quarters, nickles and dimes) for any given value?
96391
Maximize the nickels (since they have the smallest value, it takes the most of them to make up the value you want) and use 1 each of nickels and dimes.\n\neg. $1.00 = .25 + .10 + .65\n         1 quarter, 1 dime, 13 nickels\n\n\n$15,348.30= .25 + .10 + 15,347.95\n         1 quarter, 1 dime, 306,959 nickels :)\n\nOr, for a formula, subtract $0.35 (the value of 1 quarter and 1 nickel) from the amount, then divide by $0.05.  This is the number of nickels you need, add 2 to get the total number of coins.
612486
This is pretty straighforward in C - I won't write the assignment for you, but the general proccess is as follows:  \n\n1.  Count the number of each coin - store that number in some variable (say, int num_dimes;  )  If you're reading from a file, it's a little more complicated, but if your user is entering something from a command line (as is the case with beginning programming assignments, usualy) it's just a matter of having them enter a number and using a function like get() to read the input.  \n\n2.  Multiply the number of each coins by the appropriate denomination.  For example - if you counted 6 dimes in the previous step, multiply 6 by .10.  Make sure that you store the result in a floating point variable, and it's generaly a good idea to ensure that you're multiplying two floats together in order to get a float answer.  Use something like this:\n  \nfloat dimes_total; \ndimes_total = (float)num_dimes * 0.10;   \n\nthe (float) in front of num_dimes casts that variable as a floating point number.  \n\n3.  After you've done this for each coin, simply add up the totals: \n\nfloat total; \ntotal = pennies_total + nickles_total  ... etc, etc.\n\n4.  print your answer to the screen - since dollars and cents use the same number base as arithmetic done by the compiler, nothing fancy is needed here.    In cases where you're dealing with different number bases, however, it gets to be a little more tricky.
865487
3 dimes\n30 pennies\n1 quarter and 5 pennies\n2 dimes and 10 pennies\n1 dime and 20 pennies
896257
1. \nLet n be the number of nickels\nLet d be the number of dimes\nLet q be the number of quarters\n\nNumber of coins is 30:\n#1) n + d + q = 30\n\nThere are twice as many nickels as quarters:\n#2) n = 2q\n\nTotal value of the coins is $3.35.  (Use the value in cents, rather than dollars).\n#3) 5n + 10d + 25q = 335\n\nThree equations and three unknowns...\n\nSubstitute #2 into #1:\nd + 3q = 30\nAnd into #3:\n10d + 35q = 335\n\nNow solve for d:\nd = 30 - 3q\n\nSubstitute into the last one:\n10(30 - 3q) + 35q = 335\n300 - 30q + 35q = 335\n5q = 35\nq = 7\n\nPlug q back into d = 30 - 3q:\nd = 30 - 3(7)\nd = 9\n\nPlug q back into n = 2q\nn = 2(7)\nn = 14\n\nSo you have 14 nickels, 9 dimes, 7 quarters\n\n(As a double check --> 0.70 + 0.90 + 1.75 = 3.35)\n\n2. \nKaren is Anthony (statement #3)\nSince Karen is Anthony, John must be Juliet (statement #1)\nBeth is not Cleopatra, so she must be Romeo (statement #4)\nThat leaves Dan to be Cleopatra.\n\nFinal result:\nKaren = Anthony\nJohn = Juliet\nBeth = Romeo\nDan = Cleopatra\n\nThose cross-dressing friends...
927660
Hi, \n\nThey way to do this problem is to be careful in the way you set up your equations. \n\nD= dimes\nQ= Quarters\nN = Nickels\n\nSince there are twice as many dimes as quarters, you'd have to multiply the number of quarters by 2 to get it to equal the number of dimes. \n\n2Q = D \n\nSimilarly, if there are two more nickels than quarters, you'd have to add 2 to the number of quarters to get the number of Nickels\n\nQ + 2 = N\n\nNow, The total value (the number of each coin times the value of the coin.) is $7.60\n\n(0.25)Q + (0.1)D + (0.05)N = $7.60\n\nNow you have three equations with three unknowns. Now all you have to do is isolate variables, and substitute. \n\nHope that helps, \nMatt
813640
Think of this as an algebra problem.  Let's say N represents the number of quarters.  Therefore, the number of pennies = N+1 and the number of nickels = N+5.  From this you can get the equation\n\nN(.25) + (N+1)(0.01) + (N+5)(0.05) = 3.08\n\nMultiplying it out you get\n\n.25N + .01N + .01 + .05N + .25 = 3.08\n\nCombining the common values gives you\n\n.31N + .29 = 3.08\n\nFrom there you find that N is equal to 9 and the rest is easy.
511710
It depends on the coin condition.  There are many books on grading coins which you could find at your local library, and many online price guides for collectible coins:
809176
first, you need to add up 1/2, 1/4 and 1/5...first by finding a common denomonator.\n\n10/20 + 5/20 + 4/20 = 19/20\n\nSo, the first three grandkids get 19/20ths of the collection.  Meaning that the 4th grandchild gets 1/20th\n\nSo, since we know the fourth gets 49, this means that 49 = 1/20\n\nnow, cross multiply to find out the total number of coins  1/20 = 49/x\n20 x 49 =  980 (in cross multiplication, you then divide it by the remaininng number, but in this case, it's a one, so the answer will still be 980)\n\nSo, you now know the total number of coins\n\nand just need to calculate the following (I'll leave you to do it)\n\n1/2 of 980\n1/4 of 980\n1/5 of 980\n\nGood luck.
669176
p=pennies, n=nickels, d=dimes\n3p= 3,  2p+1n=7,  2p+1d=12\n\n3n=15,  2n+1p=11,  2n+1d=20\n\n3d=30,  2d+1p=21,  2d+1n=25\n\n1p+1n+1d=16\nSorry, I could only come up with 10.  :)
235611
8388608 would be the last value.\n16777215 would be the total value in cent.\n\n= $167772.15
380874
Or 7:3 if the number of coins
703550
Well, I do not know all of the world, but two American quarters are currently worth over $100,000 each in the highest grade. They are the Barber 1901 S and the Standing Liberty 1916. Also, the overdate 1918 is also that high as well. Those are the most expensive silver coins from the 20th century I know of. If it can be nickel, the 1913 Liberty head is worth over 1 million, and the 1933 Gold Double Eagle sold for about 8 million I believe.
877044
$1.19 - 3 quarters, 4 nickels, 4 pennies\n\nThe people above me who have changed there answers are cheats and should be reported as such !
223331
I'd say between $.24 and $.26.\n\nIf you find a collecter, probably a buck.
123028
13 nickels\n5 pennies\n3 dimes
982570
13 dimes, 5 quarters.  You know that there has to be an odd number of quarters because the amount ends in a 5 and there are no nickels.  Since the number is only $2.55, you know there's probably only 3 or 5 quarters, and 5 works.
789401
Let's see.\n\nTwo dimes and one nickel\nOne dime and three nickels\nFive nickels.\n\nYou can have five pennies in place of one nickel in any of the above sitiuations.\n\nSo you could have:\n\nTwo dimes and five pennies\nOne dime and two nickels and five pennies\nFour nickels and five pennies.\n\nBut once again, you can have five pennies in place of one nickel in any of the above sitiuations. (Now, D = dime, N = nickel, P = penny)\n\n1D, 1N, 10 P\n3N, 10P\n\nOne more time\n\n1D, 15P\n2N, 15P\n\nAnd again:\n\n1N, 20P\n\nOne last time\n\n25P\n\nI count 12 ways.