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Fine mathematics
This is an interesting question, which belongs to the category of indefinite equations.

First of all, we must make sure that "the same Chinese characters represent the same numbers", but there is no restriction that different Chinese characters should be different.

Numbers, as long as these numbers are not among the men with problems; Secondly, it can be determined that each Chinese character cannot be 0, otherwise it will not be called.

Two digits, only 1 ~ 9.

B, J, L and Y stand for North, Beijing, Nepal and English respectively, and H and N stand for Beijing Huanhe.

"you"

Then, Beibei =10b+b =1b, Jingjing = 1 1J, Nini = 1 1L, Yingying =/kloc-.

Then (100h+10y+n)/2008 =11b/(1165438).

Analysis:

Let (100h+10y+n)/2008 = k = b/(j-l-y)

Obviously, k must be greater than 4, otherwise, Beijing will only welcome you with four digits.

K must be less than B, that is, less than 9, because it is the quotient after B is divisible.

That is, k can only be 5, 6, 7, 8, 9.

Because b is an integer multiple of k and less than 10, it must be b = k = 5 ~ 9.

So J-L-Y= 1, J=L+Y+ 1.

When k=5,100h+10y+n =10040, H= 100, Y=4, N=0, (here, scenery, huan, especially all are 0, if they don't match,

Can be excluded); At this time, B=5, J=L+Y+ 1=L+5, and there are four combinations: L= 1, 2,3,4, j = 6,7,8,9. In ...

, Y=4, so excluding L=4, there are only three combinations: L= 1, 2,3, j = 6,7,8.

When k=6,100h+10y+n =12048, H= 120, Y=4, n = 8;; At this time, B=6, J=L+Y+ 1=L+5, and there are four combinations: L= 1, 2,3,4, j = 6,7,8,9. Where Y=4, so L=4 and B=6 are excluded, so J=6 is excluded, leaving only two combinations: L=2, 3 and J=7, 8.

When k=7,100h+10y+n =14056, H= 140, Y=5, n = 6;; At this time, B=7, J=L+Y+ 1=L+6, and there are three combinations: L= 1, 2,3, and j = 7,8,9. Where B=7, so J=7 is excluded, leaving only two combinations: L=2, 3 and J=8, 9.

When k=8,100h+10y+n =16064, H= 160, Y=6, n = 4;; At this time, B=8, J=L+Y+ 1=L+7, and there are two combinations: L= 1, 2, and j = 8,9. Where B=8, so J=8 is excluded, leaving only one combination: L=2 and J=9.

When k=9,100h+10y+n =18072, H= 180, Y=7, n = 2;; At this time, B=9, J=L+Y+ 1=L+8, and there is only one combination: L= 1, J=9. Because both b and j are 9, this example is excluded.

To sum up, * * has 3+2+2+ 1=8 answers that meet the meaning of the question. The list is as follows:

Beibei Jingjing Nini welcomes Beijing and welcomes you.

1 5566 1 144 10040

2 557722 44 10040

3 5588 33 44 10040

4 66772244 12048

5 66883344 12048

6 778822 55 14056

7 77993355 14056

8 889922 66 16064