Why are all 4-digit Fredholl numbers composite?

Fredholl numbers are number, which have 2 different digits, equal number of each one, and zero cannot be the first digit. Some examples of fredholl numbers are: 355533 and 4141.
233233, 535351 or 055050 are not fredholl numbers.

In case of 4 digit numbers let’s look at the possible forms of Fredholl numbers:

They have to be one of the following: xxyy, xyxy, xyyx
We can write above numbers as:
$$xxyy= x*1100 + y*11 = x*11*100 + y*11 = 11 (100x+y) xyxy=\\= x*1010 + y*101 = x*101*10 + y*101 =\\= 101 (10x+y) xyyx=x*1001 + y*110 = x*11*91 + y*11*10 =\\= 11 (91x+10y)$$

All 4-digit Fredholl numbers are composite because they will be divisible by either 11 or 101.

yxyx is the same case as xyxy. yyxx would be the same as xxyy, and yxxy is the same as xyyx. In all cases the first digit’s domain is {1,2.9}, and second’s digit domain is {0,1,9}, it doesn’t matter if first is called x or y, they have the same proprieties.

I don’t understand. Why do you multiply x by 1100 or whatever and add y multiplied by 11? Someone please explain! xx

Ok let me explain. Number xxyy is for example 1188, 5500, 8811. I’ll explain it on 1188. You can write 1188 as 1*1000+1*100+8*10+8. But 1*1000+1*100+8*10+8 can also be written as 11*100+11*8=11(100+8). This means that 1188 is not prime, because it can also be divided by 11. This is the case for every 4-digit fredholl number. 9922=11(900+2), 5500=11(55+0) and so on.

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