Constructing Integers with 3,4,8,9 up to 101

[The Bob]

I left it to Bob...:tongue2:

This is all I can get: $$67 = (8\times9) - \sqrt{4} -3$$

I am rubbish at this. Sorry. I will think some more.

The Bob (2004 ©)

 

[ceptimus]

There are two solutions (I know of) for 67.

One is very hard, and uses two recurring decimals. The other uses the decimal point but no recurring decimals.

I'll post them on New Year's Eve if no one gets them before and no one tells me not to.

 

[dextercioby]

$$67=\frac{8}{0.(3)|}-4.(9)$$
,where "|" is a (perfectly reflectant) mirror,and the multiplicative dot '$\cdot$' is understood.

Daniel.

PS.Ceptimus,u didn't say anything about "mirrors",right?It has 2 recurring decimals and adding the image,three in total. :tongue2:

 

[Rogerio]

There are two solutions (I know of) for 67.

One of them is (probably) :

$$67 = \frac{9}{.3 \times .4} - 8$$

... and the other one?

 

[Rogerio]

...what about 102 ? Is there a way, ceptimus ?

 

[Rogerio]

... and the other one?

Well, I think it could be

$$65 = ( 8 - .\overline{3} - .\overline{4} ) \times 9$$

OOOPS ! it should be 67 !

but... and 102 ?

 

[ceptimus]

I don't have answers for 102 or 109. All the others up to 120 I have. I've not searched beyond 120.

Well done on the 67 solution. I won't reveal the other (harder) 67 answer until New Year's Eve, just in case someone is working on it.

 

[Rogerio]

I've not searched beyond 120...

Then add 121 to the collection :-)

$$121 = \frac{8}{.4 - .\overline{3}} + .\overline{9}$$

 

[ceptimus]

Here's the difficult 67 as promised:

$$67 = \frac{.3 + .\overline{4}}{.9 - .\overline{8}}$$

Happy New Year.