數學問題......急....

2178

假設4位數=dcba

abcd=4(dcba)

abc2=4(2cba)

8bca=4(2cb8)


然後考慮兩邊百位數及十位數:
100*b+10*c=400*c+10*(4b+3)
60b-30=390c
要b跟c都是整數, 只有一個可能:
b=7
c=1

所以dcba=2178; abcd=8712

設abcd=4位數
abcd*4=dcba
dcba/4=abcd
a+a+a+a=x
a=x
d=...

2007-06-07 09:18:19 補充：
a=1or2d=7or8c=1-9b=12154no2178yes

4 (1000a + 100b + 10c +d) = 1000d + 100c + 10b +a, where 1≦a, d≦9, and 0≦a, d≦9, and they are all integers.
3999a + 390b – 60c – 996d = 0
1333a + 130b – 20c – 332d = 0
Then substitution of numbers is needed.