let x = the number of lemons
let y = the number of limes
$0.52x + $0.21y = $3.75
52x + 21y = 375
Write 52 and 375 in terms of their nearest multiple of 21
So we write 52 as (42+10) and 375 as 378 - 3
(42 + 10)x + 21y = 378 - 3
42x + 10x + 21y = 378 - 3
Divide through by 21
2x + + y = 18 -
Isolate fractions on the left:
+ = 18 - 2x - y
The right side is an integer and the left side is positive, so
both sides equals some positive integer A
+ = A and 18 - 2x - y = A
Clear of fractions:
10x + 3 = 21A
Write 21 in terms of its nearest multiple of 10
10x + 3 = (20 + 1)A
10x + 3 = 20A + A
Divide through by 10
x + = 2A +
Isolate fractions on the left
- = 2A - x
The right side is an integer, so the left side is too, say B
- = B and 2A - x = B
Clear of fractions:
3 - A = 10B
-A = -3 - 10B
A = 3 + 10B
A = 10B + 3
2A - x = B
2(10B + 3) - x = B
20B + 6 - x = B
19B + 6 = x
and since
18 - 2x - y = A,
18 - 2(19B + 6) - y = 10B + 3
18 - 38B - 12 - y = 10B + 3
-y = 48B - 3
y = -48B + 3
y = 3 - 48B
we must buy more than -1 but less than 18 limes
-1 < y < 18
-1 < 3 - 48B < 18
-4 < -48B < 18
Divide through by -48 which reverses inequalities:
> B >
> B >
Since B is an integer then B = 0 because 0 is
the only integer between those values.
Therefore since
y = 3 - 48B
y = 3 - 48(0)
y = 3 - 0
y = 3
and since
19B + 6 = x
19(0) + 6 = x
0 + 6 = x
6 = x
Therefore 6 lemons and 3 limes is the only possibility.
Edwin