3. Newton's Problem of the Fields and Cows

3. Newton’s Problem of the Fields and Cows
The following interesting problem is found in ewton s Arithmetica universalis
(1707):
a cows graze b fields bare in c days,
a cows graze b fields bare in c days,
a cows graze b fields bare in c days.
What relation exists between the 9 (positive integer) quantities a, b, c, . . . , a , b , c
It is assumed that all the fields provide the same amount of grass, that the daily growth
of the fields remains constant, and that all the cows eat the same amount each day. Let M
be the initial amount of grass in each field, m the daily growth of each field, and Q the daily
consumption of each cow.
On the evening of the first day the amount of grass remaining in each of b fields is
and on the evening of the second day
and on the evening of the third day
etc. On the evening of the c th day
bM bm aQ
bM 2bm 2aQ
bM 3bm 3aQ
bM cbm caQ
remains. Since b (b and b ) fields are grazed bare in c (c and c ) days,
1 bM cbm caQ 0,
2 b M c b m c a Q 0,
3 b M c b m c a Q 0.
Considering 1 and 2 as linear equations for unknowns M and m, we get
M cc ab ba
bb c c
Q, m a bc ab c
bb c c
[Exercise: show that if c c , then 2 is equivalent to 1, i.e., 2 is redundant. Thus to
avoid redundancy, assume that c c , say c c. ] Substitute these values into 3 and
multiply through by bbc c to get
b cc ab ba c b a bc ab c c a bb c c.
is more easily derived by noting that 1, 2, 3 is a linear homogeneous system of
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equations in three unknowns, M, m and Q with a non-trivial solution, and thus the
determinant
or
b cb ca
b c b c a 0,
b c b c a
b cb ca
b c b c a 0,
b c b c a
and this yields .
by itself is a necessary, but not sufficient condition. For example
a b c 2 3 2
a b c
a b c
a b c
a b c
a b c
3 5 2
5 8 2
4 2 3
3 3 4
2 5 5
satisfies , but leads to Q 0, i.e., cows eat nothing.
also satisfies , but leads to
m 2Q 0, i.e., m and Q are both zero, or one is negative and the other is positive.
a b c
a b c
a b c
78 10 5
117 20 8
23 5 13
satisfies , since
b cb ca
b c b c a
b c b c a
10 50 390
20 160 936
5 65 299
is row equivalent to
10 50 390
0 600 1560
0 0 0
from which we can conclude that M 26Q and m 2. 6Q with Q being any (positive)
number.
The conditions on a, b, c, . . . , a , b and c for there to be positive solutions M, m and Q are
1. a, b, c, . . . , a , b and c are positive integers with
2. c c ,
3. ab a b 0,
4. a bc ab c 0, and
5. .
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