Can You Solve the Four Nines Puzzle?

[Daminc]

No, smaller than .0000001 as well. In fact smaller than any number that is more than 0.

But still more than zero. That's the problem with using infinities. They are not real numbers just concepts. As soon as you define the number the sum can be completed.

 

[Jimmy Snyder]

But still more than zero.

If it is smaller than any number that is more than zero, and it is itself more than zero, doesn't that mean that it is smaller than itself?

 

[Greg825]

Let 0.9999999999... = x

Therefore, we have
1. 10x = 9.99999999999...
2. 10x = 9 + 0.9999999999...
3. 10x = 9 + x
4. 9x = 9
5. x = 1

x = 0.9999999999999... = 1

I numbered the steps for reference. My knowelege of what I'm noting is limited but in lines 2 and 3, can you make the substitution of .9999... with x after the multiplication of 10?

You obviously couldn't if it wasn't repeating:
x=.99
10x=9.9
10x=9+.9
.9!=x
but it is, which is why I'm posing a question rather than making a claim.

edit: I guess you could justify it by saying that as the number of 9's approaches infinity, the difference between the number and x approaches 0. So is infinity-1=infinity? (so says my calculator

 

[LarrrSDonald]

A more formal proof would be to sum up all 9/10+9/100+9/1000... per

Sum (i=1 -> inf) 9/10^i =
Lim (j -> inf) Sum (i=1 -> j) 9/10^i =
Lim (j -> inf) .9(1-10^-(j+1))/(1-1/10)=
.9/(9/10)=
1

I suppose it's as valid as anything to question if 10^-inf=1/(10^inf)=0, but they're generally pretty darn accepted as far as I've seen. One could probably further deconstruct the real number system in general, but that's beyond me - I'm a mere programmer with a side side interest in math and not much of a wizard.

One could infomally note things like that any repeating digit is equal to itself divided by 9, such as

.5555...=5/9
.123123123123...=123/999

thus perhaps venturing that

.99999...=9/9

(This is also highly useful for finding fractions of repeating numbers you bump into, .0.2857142857..=285714/999999. GCD per Euclid and divide 'em, 2/7. Bam! Ok, I'm drifting off topic. My appologies. Ok, a quick reminder, if you don't want .999.. to be 1, then it doesn't work for that particular case, only all others.)

You could also try adding up two other fractions of nine

1/9=.111111...
+
8/9=.888888...
=
9/9=.999999...

All these informal things, of course, rely on taking normal operations for correct, though personally I'm usually cool with that.

Regarding the original puzzler, I'm toying with it a bit but don't hold back on the solution on my account. This is also my first post (hence the postcount 1) so I hope I didn't embarress myself too much.

 

[Daminc]

ok, ok I'll succumb to the overwhelming attack of logic to prevail against my common sense :)

 

[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