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-Ta lập được các số
2379 a b ¯ ; 23 a b 79 ¯ ; 79 a b 23 ¯ ; 7923 a b ¯ ; a b 2379 ¯ ; a b 7923 ¯ nên số vạn của tổng cũng như số chục hay số đơn vị đều là: ( 23 +79 + a b ) x 2
- Lập luận được tổng là : (23 + 79 + a b ) x 20 202
- Tìm được : 23 + 79 + a b = 2 989 896 : 20 202 = 148
- Tính được a b = 148 – 23 – 79 = 46
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Ta lập được các số
2379ab;23ab79;79ab23;7923ab;ab2379;ab7923.
Nên số vạn của tổng cũng như số chục hay số đơn vị đều là:
(23+79+ab)x2
- Tổng là : (23 + 79 +ab ) x 2 x (10000+100+1)
=(23+79+ab) x 20202
- Vì tổng là 289896 \Rightarrow (23 + 79 + ab)x 20202= 2 989 896
\Rightarrow23+79+ab= 148
\Rightarrow ab= 148 – 23 – 79 = 46
Gọi số cần tìm là x
Theo thứ tự:
1: x- 23- 79
2: x-79-23
3:79-x-23
4: 23-x-79
5: 23-79-x
6: 79-23-x
Mà tổng tất cả là 2989896
Điều kiện:
-dù đổi vị trí ở đâu nhưng giá trị của tổng các chữ số đều bằng nhau
( tổng các chữ số ở 1, 2, 3, 4, 5, 6 đều bằng nhau)
- Tổng tất cả các số là 28989896
=>(23 + 79 +x)x2
Nhờ đó ta sẽ có tổng như sau:
[(23+79+x)x2].10000+[(23+79+x)x2].100+[(23+79+x)x2]=[(23+79+x)x2].20202
= 23+79+x=2989896 : 20202 = 148
= >x=148 - 23 - 79
= 46
ĐS: x = 46