Using the three conditions, the smallest number of horses that can be shared among the generals is 41
The conditions are given as:
The numbers that can divide 5 and leave a remainder of 1 are:
11, 16, 21, 26, 31, 36, 41, 56, 61, 66, 71......
Of all these numbers, the smallest number that satisfy the three conditions is: 41
i.e.
41/3 = 13 R 2
41/5 = 8 R 1
41/7 = 5 R 6
Hence, the smallest number of horses is 41
Let the number of horses a general gets be x.
The explicit rule for each condition can be represented as:
Using the condition (b), we have:
Total = 5x + 1
Rewrite as a function
f(x) = 5x + 1
When x = 0, we have:
f(0) = 1
For other values of x, we have:
f(1) = 6
f(2) = 11
f(3) = 16
Rewrite as:
f(0) = 1
f(1) = 5 + 1 = 5 +f(0)
f(2) = 5 + 6 = 5 + f(1)
f(3) = 5 + 11 = 5 + f(2)
Express 2 as 3 - 1
f(3) = 5 + f(3 - 1)
Express 3 as x
f(x) = 5 + f(x - 1)
Hence, the recursive function is f(x) = 5 + f(x - 1) where f(0) = 1
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