Arabic Problem

This is an old arabic problem:

An old man had 11 horses. When he died, his will stated the following distribution to his 3 sons:
1/2 gives to the eldest son,
1/4 for 2nd son,
1/6 for 3rd son.

Find: how many horses each son gets ?

There are 2 methods to solve: first using simple arithmetic trick without knowing the theory behind; the second method will explain the first method “from an advanced standpoint” – Number Theory (Felix Klein’s Vision )

1) Arithmetic trick:

11 is odd, not divisible by 2, 4 and 6.

Loan 1 horse to the old man:
11+1 = 12

1st son gets: 12/2 = 6 horses
2nd son gets:12/4 = 3 horses
3rd son gets: 12/6 = 2 horses

Total = 6+3+2=11 horses

Up to you if you want the old man to return the 1 loan horse 🙂

Strange! WHY ?

2) From an advanced standpoint: Number Theory

The mystery is unveiled here:

\boxed{\displaystyle\frac{11}{12} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} }
2|12, 4|12, 6|12

Let’s generalize the Arabic problem: old man has n horses, given to 3 sons x, y, z horses, respectively, such that:
x + y + z = n
and
x |n+1, y |n+1, z |n+1
Without loss of generality, we assume
x > y > z

Let n+1 = ax = by = cz
for a, b, c integers

From x + y + z = n, we obtain:
\displaystyle \frac{n+1}{a} + \frac{n+1}{b} + \frac{n+1}{c} = n

\boxed{\displaystyle\frac{n}{n+1} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} …….[1]
a | n+1, b | n+1, c | n+1, a < b < c

Note: x > y > z
(n+1)/a > (n+1) / b > (n+1) / c
n > 0
1/a > 1/b > 1/c
a < b < c

For n > 0,
\frac{ n}{n+1} < 1 \implies a \geq 2

Applying unique prime factorization from
Fundamental Theorem of Arithmetic (FTA)

[Note: we can prove using this FTA theorem that a > 2 is impossible !]

Let a = 2, substitute in [1]:

\frac{1}{b} + \frac{1}{c} = \frac{n}{n+1} -  \frac{1}{2} = \frac{1}{2}- \frac{1}{n+1}

\boxed {\displaystyle\frac{1}{b} + \frac{1}{c} = \frac{1}{2}- \frac{1}{n+1} }

We can deduce (omit the manual FTA deduction or computer steps by trials and eliminations): n has 6 possible values (7, 11, 17, 19, 23, 41) and 7 ways of distributions (a, b, c) as below:

n =7, a=2, b=4, c=8
n =11, a=2, b=4, c=6
n =11, a=2, b=3, c=12
n =17, a=2, b=3, c=9
n =19, a=2, b=4, c=5
n =23, a=2, b=3, c=8
n =41, a=2, b=3, c=7

For the Arabic problem:
n =11, a=2, b=4, c=6

ax = n+1 => 2x =12 => x = 6 horses
by = n+1 => 4y =12 => y = 3 horses
cz = n+1 => 6z =12 => z = 2 horses

Verify:
n = x + y + z = 6 + 3 + 2 = 11 horses

\boxed{\displaystyle\frac{11}{12} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} }
2|12, 4|12, 6|12

[Reference] "单位分数" 柯召 ,孙琦 -北京:科学出版社,2002,数学小丛书(14)

Note: Prof Ke Zhao 柯召 (1910-2002) had Erdös # 1 with a Theorem Erdös-Ke-Rato in Combinatorics.

http://www.hlhl.org.cn/english/showsub.asp?id=679

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