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:

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:

…….[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,

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]:

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**

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

Reblogged this on Singapore Maths Tuition.