Round 167
The 'Lenny' authors, in Round 167, wrote:Fifteen different farmers each had their own apple orchards. The first farmer had only one apple tree in his orchard, the second farmer had two trees in his orchard, and so on up to fifteen trees for the fifteenth farmer.
At harvest time, each farmer noticed something very peculiar: each tree in the same orchard produced the exact same number of apples. (Note that the number of apples per tree varied from orchard to orchard) And when all fifteen farmers got together to talk about it, they all realized that, if the farmer with eleven trees had given one apple to the farmer with seven trees, and the farmer with fourteen trees had given three apples each to the farmer with nine trees and the farmer with thirteen trees, they'd all end up with the exact same number of apples!
How many apples did all the farmers produce, in total?
Define a, b, c, d, and e as follows:
. . .a: number of apples per tree in 7th orchard
. . .b: number of apples per tree in 9th orchard
. . .c: number of apples per tree in 11th orchard
. . .d: number of apples per tree in 13th orchard
. . .e: number of apples per tree in 14th orchard
Then we have:
. . .11c - 1 = 7a + 1
. . .14e - 6 = 9b + 3
. . .14e - 6 = 13d + 3
. . .. . .7a - 11c = -2
. . .9b - 14e = -9
. . .13d - 14e = -9
. . .11c - 14e = -5
Adding the first and fourth equations, we get the following reduced system:
. . .7a - 14e = -7
. . .9b - 14e = -9
. . .13d - 14e = -9
Note that the last two equations require that 9b = 13d.
Subtracting the third equation from the second and first equations, we further reduce the system and get:
. . .7a - 13d = 2
. . .9b - 13d = 0
Solving the above (under-determined) system, we get:
. . .a = (13/7)d + 2/7
. . .b = (13/9)d
For this to be a sensible system, we must have the right-hand sides come out to whole numbers. This confirms that d is a multiple of 9 and that b is a multiple of 13. Start testing numbers with the first multiple of 9:
. . .. . .d = 9: (13/9)d = 13 = b
Also, if b = 13, then there are 9(13) + 3 = 117 + 3 = 120 apples in each orchard (after the farmers make their adjustments). So try "d = 9" as a solution for the entire system. If d = 9, then:
. . .a = (13/7)(9) + 2/7 = 117/7 + 2/7 = 119/7 = 17
. . .b = 13
Since 7a - 11c = -2, then 11c = 7a + 2, and:
. . .c = (7a + 2)/11 = (119 + 2)/11 = 121/11 = 11
. . .d = 9 (by assumption)
Since 11c - 14e = -5, then 14e = 11c + 5, and:
. . .e = (11c + 5)/14 = (121 + 5)/14 = 126/14 = 9
Therefore, we have:
. . .farmer 7 has 7(17) + 1 = 119 + 1 = 120 apples
. . .farmer 9 has 9(13) + 3 = 117 + 3 = 120 apples
. . .farmer 11 has 11(11) - 1 = 121 - 1 = 120 apples
. . .farmer 13 has 13(9) + 3 = 117 + 3 = 120 apples
. . .farmer 14 has 14(9) - 6 = 126 - 6 = 120 apples
So the solution checks.
However, this is, as was noted earlier, an underdetermined system. The generalized solution is of the form:
. . .(a, b, c, d, e) = (2e - 1, (14/9)e - 1, (14/11)e - 5/11, (14/13)e - 9/13, e)
That is, any value of e which returns whole-number values for a, b, c, and d, and for which 14e - 6 is a multiple of 120, will work. For instance:
. . .e = 25749
. . .a = 51497
. . .b = 40053
. . .c = 32771
. . .d = 27729
This yields a per-orchard value of 360480 = 3004(120), and a total value of 5407200.
It is reasonable to assume, however, that the smaller total value of 1800 is the expected answer.
Eliz.
Well i got 1800 BUT at school i failed maths soooo 
