There are TWO solutions.
There's a coin collection of pennies, nickels, dimes, and quarters worth $10.14 altogether.
(1) p + 5n + 10d + 25q = 1014
The pennies and nickels together are worth $1.09,
(2) p + 5n = 109
the nickels and dimes together are worth $5.15,
5n + 10d = 515
Dividing that through by 5
(3) n + 2d = 103
and the dimes and quarters together are worth $9.05.
10d + 25q = 905
Dividing that through by 5
(4) 2d + 5q = 181
Writing every number in (4) in terms of the nearest multiple
of the coefficient of the term whose absolute value is
smallest, which is 2:
2d + 4q + q = 180 + 1
Dividing through by 2
d + 2q + q/2 = 90 + 1/2
Isolate fractions:
q/2 - 1/2 = 90 - d - 2q
The right side is an integer, and so is the left side.
The left side is non-negative, so let that non-integer
integer be A:
(5) q/2 - 1/2 = A
(6) 90 - d - 2q = A
Multiply both sides of (5) by 2
q - 1 = 2A
(7) q = 2A + 1
Substitute in (6)
90 - d - 2q = A
90 - d - 2(2A + 1) = A
90 - d - 4A - 2 = A
88 - 5A = d
(8) d = 88 - 5A
Substitute in (3)
n + 2d = 103
n + 2(88 - 5A) = 103
n + 176 - 10A = 103
(9) n = 10A - 73
Substitute in (2)
p + 5n = 109
p + 5(10A - 73) = 109
p + 50A - 365 = 109
(10) p = 474 - 50A
Since n > 0, using (9),
10A - 73 > 0
10A > 73
(11) A > 7.3
Since p > 0, using (10)
474 - 50A > 0
-50A > -474
(12) A < 9.48
So from (11) and (12), A is either 8 or 9
So it appears that we have two solutions:
Using A = 8,
from (10), p = 474 - 50A = 474 - 50(8) = 474-400 = 74
from (9), n = 10A - 73 = 10(8) - 73 = 80 - 73 = 7
from (8), d = 88 - 5A = 88 - 5(8) = 88 - 40 = 48
from (7), q = 2A + 1 = 2(8) + 1 = 16 + 1 = 17
So the first solution is 74 pennies, 7 nickels, 48 dimes, 74 pennies
Using A = 9,
from (10), p = 474 - 50A = 474 - 50(9) = 474-450 = 24
from (9), n = 10A - 73 = 10(9) - 73 = 90 - 73 = 17
from (8), d = 88 - 5A = 88 - 5(9) = 88 - 45 = 43
from (7), q = 2A + 1 = 2(9) + 1 = 18 + 1 = 19
So the second solution is 24 pennies, 17 nickels, 43 dimes, 24 pennies.
We check to see if (1) is satisfied. We didn't use it at all!
p + 5n + 10d + 25q = 1014
74 + 5(7) + 10(48) + 25(17) = 74+35+480+425 = 1014
So the first solution checks.
p + 5n + 10d + 25q = 1014
24 + 5(17) + 10(43) + 25(19) = 24+85+430+475 = 1014
So the second solution checks, too!
Edwin