Can You Solve the Four Nines Puzzle?

[LarrrSDonald]

Heh. Well, it is very counter-intuitive - you best believe I didn't want to buy it first time I was told of thie atrocity (around 10th grade, I believe). However, series and other number theory things often don't look very sensible in the slightest to me, for instance 1/(10^inf) feels like it ought to be way less then 1/inf, after all the bottom number is an infinite order of magnitude rather then just plain infinite. Still, they sneakily both converge on zero lim->inf (one faster then the other of course, when that matters). Are there more integers then even numbers? Then primes? Kinda all infinite, aren't they? It's all nuts I tells ya!

 

[Rogerio]

Some results:

$$65 = (.\bar{9} + .\bar{9}) ^ { (\sqrt{9}) ! } + .\bar{9}$$

$$67 = (.\bar{9} + .\bar{9}) ^ { (\sqrt{9}) ! } + \sqrt{9}$$

$$68 = \frac{((\sqrt{9}) !) !} {9} - 9 - \sqrt{9}$$

 

[Integral]

Please do not continue the .999... =1 discussion in this (or any other) thread. Read this

Note that it is locked. This has been discussed enough. It is a mathematical fact.

 

[Daminc]

Another thing which I'm noticing. When I was at school we used a dot above the number to imply an infiniate recurance. The examples here shows a line above the number.

When did this change?

 

[LarrrSDonald]

Think I have a full set 0-132. Unfortunatly my host is temporarily down and I'm not so sure about jammig 133 lines in this thing - that'd be mucha annoying if someone else did it.

Possible spoiler in white:
I found no trick per se, though I can't say I looked for one too hard. Mostly, my plan was to find a largish (or appropriatly sized, though I'd mostly typed out the lower ones while I was thinking) number (99, 9*9, 9!/9, 9!/(Sqrt[9]!), etc) with two 9s and then blanket the surrounding area with +-{1,3,6,9} (3=Sqrt[9], 6=3!, 1=.9..) with the other two. I'm sure this could be done more systematically then just typing them out willie nelly, but I didn't. With two nines you can blanket +-10 completely (+9+1, +9*1, +9-1, +6+1, +6*1, +6-1, +3+1, +3*1, +3-1, +1+1, +1*1 for positive, invert for neg), as well as sporadic numbers outside (9+3, 9+6, et al). This covers a ton of ground, I assume this is the 3 min to 5-6 left mentioned unless I'm missing something else. It's obviously helpful to blackbox any "single nine" operation, thinking of a nine as a {1,3,6,9} of your choice rather then a 9. The last few were just grunt work, 132=(Sqrt[9]!-.9)^Sqrt[9]+Sqrt[9]! was the final conquest.

[EDIT] Typo. Yes, I'm sure there's plenty more.

 

[Jimmy Snyder]

132=(Sqrt[9]!-.9)^Sqrt[9]+Sqrt[9]!

But 131 = (Sqrt[9]!-.9)^Sqrt[9]+Sqrt[9]!?

 

[LarrrSDonald]

Opps, my bad, 132=(Sqrt[9]! ! / Sqrt[9]!) + Sqrt[9] + 9, though 131=... per above was the last discovered. Typed 132=... when I copied it for some reason. Sorry 'bout that.

 

[LarrrSDonald]

Thought a little more about this..

As I mentioned, +-10 (inclusive) with two nines can be made with one from +9,+6,+3,*1 with -1,*1,+1 on each where 3=Sqrt[9], 6=3! and 1=.9... (Yeah, just repeating it for completeness).

Thus,

9+1 = 10 (covers 0-20)
9*3 = 27 (covers 17-37)
6*6 = 36 (covers 26-46)
9*6 = 54 (covers 44-64)

Gap: 65-70

9*9 = 81 (covers 71-91)
99 = 99 (covers 89-109)
6!/6 = 120 (covers 110-130)

Gap: 131-132

So 65-70, 131, 132 left - i.e. eight of them. Additionally, +9+3 and +9+6 w/ 54 can knock off 66 and 69 (or 81 minus the same), so 65, 67, 68, 70 for that gap. 6!/6=120 also picks up 132 w/ +9+3. 6!/9=80, so that juuust dips down to grab 70 and hits 68 w/ -9-3. The remaining three are tricky, but (1+1)^6=64 (using three 9s) and hit 65,67 with +1 and +3. 131 is covered above, but just for the hell of it:

131=(Sqrt[9]!-.9)^Sqrt[9]+Sqrt[9]!

Obviously replace 1,3,6 with appropriate single nine op when I haven't. Belive that's the lot of 'em.

 

[LarrrSDonald]

Here's a similar solution to two 4s, two 9s, 0 to 232:

With one 4 one 9, one can generate a +-13 span,

0=*(.9..)^4
1=Sqrt[4]-(.9..)
2=Sqrt[4]*(.9..)
3=4-(.9...)
4=4*(.9...)
5=4+(.9...)
6=Sqrt[9]*Sqrt[4]
7=Sqrt[9]+4
8=Sqrt[9]!+Sqrt[4]
9=Sqrt[9]^Sqrt[4]
10=Sqrt[9]!+4
11=9+Sqrt[4]
12=Sqrt[9]*4
13=9+4

Neg by -same. Note that +-0 or *1 can no longer be considered obvious (though not amazingly hard either), there is no single four op to 1 or 0. Thus, 9=9, for instance, doesn't adequitly show that you can get rid of your (in this case) extra 4.

The following ranges can then be covered (Gaps noticed rather then covered):

9+4=13 {0,26}
4*9=36 {23,49}
49 {36,62}
4!*Sqrt[9]=72 {59,85}
94 {81,107}
(4+(.9..))!=120 {107,133}
4!*(Sqrt[9]!)=144 {131,157}
-- Gap {158,166}
(Sqrt[9]!)/4=180 {167,193}
Gap {194,202}
4!*9=216 {203,229}
-- Gap {230,232}

Then,

Covering {158,166}:

9*9*Sqrt[4] w/ -4,-2,+2,+4 covers 158 160 164 166
Sqrt[9]^4*Sqrt[4] w/ -3,-1,*1,+1,+3 covers 159 161 162 163 165

Covering {194,202}:

99*Sqrt[4] w/ -4,-2,+2,+4 covers 194 196 200 202
49*4 w/ -1,+3 covers 195 199

197=94*Sqrt[4]+9
198=99*Sqrt[Sqrt[4]]^Sqrt[4]
201=(4!-Sqrt[4])*9+Sqrt[9]

Covering {230,232}:

230=(4!-(.9..))*(Sqrt[9]!+4)
231=(4!+Sqrt[4])*9-Sqrt[9]
232=(Sqrt[9]!)^(Sqrt[9])+4*4

Which should altogether cover {0,232}. As a bonus,

(4!+Sqrt[4])*9 w/ -1,*1,+1 takes the next three - 233,234,235. Sorry if this is a bit longish, kind of needed lots of lines rather then long ones..

 

[tiabobia]

got all except for 65

hmmmm i have all of them but i can't seem to get 65... either my brains not working from the thinking or I am just really stupid... or 65 is HARD to get... hmm help

 

[Spazzcat]

Stuck on 131

I started this puzzle yesterday evening, doing 1-100. I'd gotten all but 85, 88, and 94 when I came across this forum. With .9...=1, I've gotten everything except 131, because according to my calulator, the above formula does NOT equal 131.

Maybe I'm just not reading it right. Could someone explain the equation for 131?

 

[Rogerio]

I started this puzzle yesterday evening, doing 1-100. I'd gotten all but 85, 88, and 94 when I came across this forum. With .9...=1, I've gotten everything except 131, because according to my calulator, the above formula does NOT equal 131.

Maybe I'm just not reading it right. Could someone explain the equation for 131?

$$131 = { ( (\sqrt{9}) ! - .\bar{9} ) ^ \sqrt{9} } + (\sqrt{9}) !$$

or

$$131 = { ( 3! - 1 ) ^ 3 } + 3!$$

or

$$131 = { 5 ^ 3 } + 6$$