 Posers and Puzzles

1. 02 Jul '04 01:44
See how many integers you can form, starting with 1 and going through 10, by using only the digit 4 four times - no more, no less - and the multiplication, division, addition and/or subtraction signs.

Samples:

1 = 44/44
2 = 4/4 + 4/4

-Ray.
2. 02 Jul '04 12:59
3= (4+4+4)/4
4= (4-4)/4 + 4
5= ((4x4)+4)/4
6= 4+(4+4)/4
7= 4+4-4/4
8= (4+4)x4/4
9= 4+4+4/4
10=(44-4)/4
11=44/sqrt(4x4)
12=4x4-sqrt(4x4)
13=44/4+sqrt(4)
14=4x4-(4/sqrt(4))
15=44/4+4
16=4+4+4+4
17=4x4+4/4
18=4x4+4/sqrt(4)
3. 02 Jul '04 13:45
Good work!

-Ray.
4. 02 Jul '04 16:482 edits
19 = 4! - (4 + 4/4)
20 = 4! - 4 + 4 - 4
21 = 4! -4 + 4/4
22 = 4! - sqrt(4) + 4 - 4
23 = 4! - sqrt(4) + 4/4
24 = 4*4 + 4 + 4
25 = 4! + sqrt(4) - 4/4
26 = 4! + sqrt(4) + 4 - 4
27 = 4! + 4 - 4/4
28 = 4! +4 - 4 + 4
29 = 4! + 4 + 4/4
30 = 4! + sqrt(4) + sqrt(4) + sqrt(4)

and not sure this is optimal, but...
31 = 4! + 4/(.4 recurring) - sqrt(4)

EDIT
better is:
31 = 4!+ (4! + 4)/4
5. 02 Jul '04 18:56
I think that after passing 30, all mathematical symbols are acceptable. Therefore, your recurrence dot is acceptable.

Keep going! 😉

-Ray.
6. 04 Jul '04 08:44
Originally posted by rgoudie
I think that after passing 30, all mathematical symbols are acceptable.
Are also standard functions acceptable? Is it valid if I define ONE(x) as a function such that ONE(4) = 1, that could be defined with standard functions as:

floor(cubic root(x)) or
ceil(log(x)) or
sign(x)

This way,

32 = (4+4) x sqrt(4x4)
33 = 4! + 4 + 4 + ONE(4)
34 = 4! + 4 + 4 + sqrt(4)
35 = 4! + (4! / sqrt (4)) - ONE(4)
36 = 4^4 - 4! - 4

-- Germano
7. 04 Jul '04 14:56
Is Log acceptable?....
8. 04 Jul '04 15:42
If square root is acceptable, then so is log.
I'm not too sure about that one function. 🙂

-Ray.
9. 05 Jul '04 10:29
Originally posted by rgoudie
I'm not too sure about that [b]one function. 🙂

-Ray.
[/b]
it's only a way to say one of the other (valid, as you confirmed) three... 🙂 just substitute it with ceil (log(4)), this is the sense. Anyways, 36 was wrong, it's quite easily

36 = (4+4)x4 + 4

while 37... I really can't figure it! 8-)
10. 05 Jul '04 21:46
without ONE, CEIL etc...

33 = 44 - 4/(.4 recurring) - sqrt(4)
34 = 4! + 4 + 4 + sqrt(4)
35 = 44 - 4/(.4 recurring)
36 = 44 - 4 - 4
37 = 4! + 4 + 4/(.4 recurring)
38 = 44 - 4 - sqrt(4)
39 = 4!*sqrt(4) - 4/(.4 recurring)
40 = 44 - sqrt(4) - sqrt(4)

11. 06 Jul '04 23:11
If log is acceptable then there is a proof that any integer can be expressed with four fours...
12. 07 Jul '04 00:20
With a specific base or will any do?

-Ray.
13. 08 Jul '04 17:411 edit
Any old kind of log will do. By the way, is the % operation allowed? (which would mean divide by 100, e.g. 20% = 0.2)