[Guest]

This may have been posted before.

How high can you count using four fours. You can use all mathematical symbols etc. and concatenate the 4's e.g. 4444. here's a start,

44/44 =1

4/4 + 4/4 =2

(4+4+4)/4 =3

There is a record for the highest number achieved somewhere, I'm sure one of you will know where.

[Guest]

here is my approach for 107:

First I must explain my notation that I had used at my previous post:

I used 0.4^- as 0.444444444 and it's also equal to 4/9. Indeed - sign should be just over the "4"

I suppose this is a valid notation, but any objection will be accepted by me.

So:

107= 44/.4^- + 4 + 4

= 44/(4/9) + 4 + 4

= 44 * 9 /4 + 4 + 4

= 11 * 9 + 4 + 4

= 99 + 4 + 4

your math is solid. but i count 5 fours instead of 4.

[Guest]

107= 44/.4^- + 4 + 4 ----->

your math is solid. but i count 5 fours instead of 4.

Sorry, a silly mistake!

Here is a new approach:

107=4!/.4^- + 4!/.4^- - d/dx(x)

107=24/0.44444 + 24/0.44444 - d/dx(x)

107=24/(4/9) + 24/(4/9) - d/dx(x)

107=54 + 54 - 1

since you can add many d/dx(x), it's possible to write all numbers up to infinite.

but the riddle makes no sense then, so it shouldn't be valid.

but I'll wait for OP to rule out this.

[Guest]

Sorry, a silly mistake!

Here is a new approach:

107=4!/.4^- + 4!/.4^- - d/dx(x)

107=24/0.44444 + 24/0.44444 - d/dx(x)

107=24/(4/9) + 24/(4/9) - d/dx(x)

107=54 + 54 - 1

since you can add many d/dx(x), it's possible to write all numbers up to infinite.

but the riddle makes no sense then, so it shouldn't be valid.

but I'll wait for OP to rule out this.

444/4 - 4 = 111 - 4 = 107

[Guest]

444/4 - 4 = 111 - 4 = 107

an iteration of my mistake: 5 fours

[Guest]

(4!+4!-.4-)/.4-

[Guest]

Sorry, a silly mistake!

Here is a new approach:

107=4!/.4^- + 4!/.4^- - d/dx(x)

107=24/0.44444 + 24/0.44444 - d/dx(x)

107=24/(4/9) + 24/(4/9) - d/dx(x)

107=54 + 54 - 1

since you can add many d/dx(x), it's possible to write all numbers up to infinite.

but the riddle makes no sense then, so it shouldn't be valid.

but I'll wait for OP to rule out this.

You're correct, the riddle will make no sence if you use d/dx(x)

[Guest]

I think its 3.408e+38 approx

4^4 = 256

256 ^ 4 = 4294967296

4294967296 ^ 4 = 3.408 e+38

[plainglazed]

4! + 4! + (4!/.4) = 108

√4(4!/.4¯) + Γ√4 = 109

√4(4!/.4¯) + √4 = 110

44.4/.4 = 111

44/.4 + √4 = 112

4(4! + 4) + Γ√4 = 113

44/.4 + 4 = 114

(4! + 4! - √4)/.4 = 115

44/.4 + Γ4 = 116

(4! + 4! + 4)/4¯ = 117

√4(4!/.4) - √4 = 118

(4! + 4! - .4)/.4 = 119

(44 + 4)/.4 = 120

(4! + 4! + .4)/.4 = 121

[Prof. Templeton]

I've got a list of numbers created with 4 fours that goes from 1 to 40,000 with only 5 missing entrys. I downloaded it last year when this same question was posted if anyones interested. FYI.