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廢話不多說，題目如下：


某二位數除以此數的反序數(各位數字倒過來。如54的反序數是45)所得的商與餘數相同。請問此數為何？




答案是52，但我不知道為什麼，請各位大大詳細的解說一下。

廢話不多說，解說如下：


(Ⅰ)
設此二位數的十位數字為a,個位數字為b,1≦a≦9,0≦b≦9
則此數為10a+b,此數的反序數為10b+a
又設此數除以此數的反序數所得的商與餘數皆為r
則10a+b=(10b+a)r+r

(Ⅱ)
假設b為0,則10a+b=10a=(10b+a)×10+0
但是商與餘數不相同,故假設錯誤,b不為0
所以1≦b≦9

(Ⅲ)
簡化10a+b=(10b+a)r+r,求r的範圍
10a+b=(10b+a+1)r
r=(10a+b)/(10b+a+1)
r=10-(99b+10)/(10b+a+1)

因為10a+b與10b+a+1與99b+10皆為正數
所以0<r<10

(Ⅳ)
簡化10a+b=(10b+a)r+r,求a與b的關係式
10a+b=10br+ar+r
(10-r)a=(10r-1)b+r

假設r為1,則9a=9b+1,9≦9a≦81,9b+1可能是10,19,28,37,46,55,64,73
但是9b+1不可被9整除,故假設錯誤,r不為1

假設r為2,則8a=19b+2,8≦8a≦72,19b+2可能是21,40,59
只有40可被8整除,所以8a=19b+2=40,a=5,b=2

假設r為3,則7a=29b+3,7≦7a≦63,29b+3可能是32,61
但是29b+3不可被7整除,故假設錯誤,r不為3

假設r為4,則6a=39b+4,6≦6a≦54,39b+4可能是43
但是39b+4不可被6整除,故假設錯誤,r不為4

假設r為5,則5a=49b+5,5≦5a≦45
但是49b+5>45,故假設錯誤,r不為5

假設r為6,則4a=59b+6,4≦4a≦36
但是59b+6>36,故假設錯誤,r不為6

假設r為7,則3a=69b+7,3≦3a≦27
但是69b+7>27,故假設錯誤,r不為7

假設r為8,則2a=79b+8,2≦2a≦18
但是79b+8>18,故假設錯誤,r不為8

假設r為9,則a=89b+9,1≦a≦9
但是89b+9>9,故假設錯誤,r不為9

(Ⅴ)
52=25×2+2
經過驗算確定52除以25所得的商與餘數皆為2
所以此數為52

Let the original number be x=10a+b, then

x = 10a + b = q (10b +a) + q = q(10b+a+1)

where q is the quotient. Before using the brutal force method, try to extract as most informatin as possible:

1. q is not equal 0 (otherwise x =0)
2. q is not equal to 1, because if so, then 10(a-b) + (b-a) =1, which has no integral solution.
==> q >= 2
    ==> (1) b must be less than 5, otherwise 10bq would be equal to or larger than 100 (three digits)
    ==> (2) a > b, otherwise q would be zero.

3. b is not equal to 0, otherwise, the remainder would be zero.

4. a is not equal to 9, otherwise a+1=10, q(10b+a+1)=10*q(b+1), which means the b=0
   So a is less than or equal to 8.
5. Rewrite the first equation as: 10(a-bq) = q(a+1) - b
  ==> (1) Since a is larger than b, we know q(a+1)-b = 10(a-bq) would be larger than 0.
  ==> (2) So (a-bq)>=1, 8>=a>=1+bq>=1+2b, which means b is less than or equal to 3.5. If consider b as an integer, it means that b is less than 4.
  ==> (3) If (a-bq) is equal to or greater than 2, then q(a+1)-b would be larger than 20
          ==> q(a+1) >20, since q>=2
          ==> a+1 > 10, contradiction! because a is less than 9.
  
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So  we got the relation: a-bq = 1 or a=bq+1
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==> 10(bq+1) + b = q(10b+bq+2)
==> 10 + b = q(bq+2) is equal to larger than 2(2b+2) = 4b + 4 (by applying the fact that q should be equal to or larger than 2).
==> b should be equal to or less than 2.

In addition, 10 + b = q(bq+2) >= q(2+2)=4q
==> 12 >= 10+b >=4q, which means that  q is equal to or less than 3


Now it is a good timing to use the brutal force method:
For the set (a=bq+1,b,q), only four possilbe solutions ( 1<= b <=2, and 2<= q <=3)
(3,1,2), (5,2,2), (4,1,3), (7,2,3)

check each case, only x = 52 fits the requirement.

(用中文打字, 好像都會漏字, 所以只好用英文說明)