SOLUTION: I have $2.35 in nickles and dimes . If i have a total of 23 coins , how many of each coin do i have ?

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Question 251225: I have $2.35 in nickles and dimes . If i have a total of 23 coins , how many of each coin do i have ?
Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!
In solving coin problems, you have to keep track of the number of coins and the value of the coins.
n = number of nickels
5n = value of the nickels in cents
d = number of dimes
10d = value of the dimes in cents
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We are told n+d = 23 coins.
We are told that 5n + 10d = 235 cents.
.
We can solve these two equations in several ways. An obvious approach is to set n = 23-d and then substitute.
.
5(23-d) + 10d = 235
115 - 5d + 10d = 235
5d = 120
d = 24
.
n + d = 23
n = -1
.
Checking our values:
5n = 5*-1 = -5 cents
24d = 10*24 = 240 cents
Adding: 235 cents.
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But this is a nonsensical answer. You cannot have -1 nickels.
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Using a 'brute force' approach, it is easy to see that the maximum value of 23 coins occurs with 22 dimes and 1 nickel...and it is 225 cents ($2.25), which is less than the stated amount.
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n	d	n+d	5n	10d	5n+10d
1	22	23	5	220	225
2	21	23	10	210	220
3	20	23	15	200	215
4	19	23	20	190	210
5	18	23	25	180	205
6	17	23	30	170	200
7	16	23	35	160	195
8	15	23	40	150	190
9	14	23	45	140	185
10	13	23	50	130	180
11	12	23	55	120	175
12	11	23	60	110	170
13	10	23	65	100	165
14	9	23	70	90	160
15	8	23	75	80	155
16	7	23	80	70	150
17	6	23	85	60	145
18	5	23	90	50	140
19	4	23	95	40	135
20	3	23	100	30	130
21	2	23	105	20	125
22	1	23	110	10	120

.
So there is no solution to the problem.

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