~ 祝賀本戶3 周年誕辰 ~ 3 題贈3*14 點

Let the equation be ABCAB * A = DDDDDD, so
(1001 * (10A + B) + 100C) * A = 1001 * 3 * 37 * D
As C is a unit digit (0~9), it cannot be the multiple of 1001,
so it can only be 0. The equation becomes :
(10A + B) * A = 3 * 37 * D
As A =/= D, so A can only be 3.
(If D = 2, A = 6, then 60 + B is greater than 37;
if D = 2, A = 3, then 30 + B is smaller than 74).
The equation becomes :
30 + B = 37 * D
Obviously, B is 7 and D is 1.
Ans : 三 = 3, 年 = 7, 又 = 0, 等 = 1. (37037 * 3 = 111111)
As a + 1/a = b + 1/b, so
(a - b) = 1/b - 1/a = (a - b) / ab
therefore,
Case I, a = b (and also = c), the triangle is an equilateral triangle.
Case II, ab = 1 (and a > 1), so b = 1/a and c is either a or 1/a
(which must satisfy the triangle inequality b + c > a)
So the triangle is an isosceles triangle.
Since equilateral triangle is also a type of isosceles triangle, so :
Ans : It is an isosceles triangle.
Let P = k1 * k2 * . . . * k33 = 1 . . . . (given)
S1 = Sum of P divided by one's k, ie
S1 = P/k1 + P/k2 + . . . + P/k33
As AM >= GM, so S1 >= 33C32 * P^(33C32 - 32C31)
As P = 1, therefore S1 >= 33C32
Similarly,
S2 = Sum of P divided by two's k, ie
S2 = P/(k1*k2) + P/(k1*k3) + . . . + P/(k32*k33)
As AM >=GM, so S2 >= (33C31) * P^(33C31 - 32C30)
As P = 1, therefore S2 >= 33C31
. . .
Sn >= 33C(33-n) * P^[33C(33-n) - 32C(32 - n)] = 33C(33-n)
. . .
S33 >= 33C0 * P^[(33C0 - 32C0) = 33C0
So,
(2 + k1)*(2 + k2)*(2 + k3)* . . . *(2 + k33)
= 2^33 + S32 * 2^32 + S31 * 2^31 + ... + Sn * 2^n + ... + S1 * 2 + P
>= 2^33 + 33C1 * 2^32 + 33C2 * 2^31 + ... + 33C(33-n) * 2^n + ... + 33C32 * 2 + 1
= (2 + 1)^33
= 3^33
Therefore,
(2 + k1)*(2 + k2)*(2 + k3)* . . . *(2 + k33) >= 3^33

(1) 37037X3=111111

(1)

考慮 111111 = 3 × 7 × 11 × 13 × 37 , 三年又三年 要整除於 11，則 三 + 又 + 年 = 年 + 三 , 又 = 0 ,

三年 0 三年 = 10010 三 + 1001年 = 1001(10三+年) = (7)(11)(13)(10三+年) , 則三不整除於 7，整除於 3，

2011-12-29 19:39:30 補充：
若 三 = 3 , (40000)(3) = 120000 > (3年03年)(3) > (30000)(3) = 90000 , 3年03年 = 111111 , 年 = 7 .

若 三 = 6 , 444444 > (70000)(6) = 420000 > (6年06年)(6) > (60000)(6) = 360000 > 333333 不合 .

若 三 = 9 , (100000)(9) = 900000 > (9年09年)(9) > (90000)(9) = 810000 , 9年09年 = 888888 , 年 = 798 不合 .

2011-12-29 19:39:34 補充：
原算式 =

37037
×___ 3
-----------
111111

2011-12-29 19:54:57 補充：
(2)

a + 1/a = b + 1/b
(ab)(a+1/a) = (ab)(b+1/b)
a²b + b = ab² + a
a²b - ab² = a - b
ab(a-b) = 1(a-b)
(ab-1)(a-b) = 0
a = b or ab = 1
a = b or 1/b

當 a = 1/b , a = 2 , b = 0.5 , c = 2 時則不是等邊三角形 .