Can You Solve the Four Nines Puzzle?

[ACG]

Hi! I have a puzzle here and was wondering if you guys could solve it.

Using exactly four nines each time and no other numbers or letters, find expressions which evaluate to every integer between 0 and 132. You may put parentheses wherever you wish and uses whatever mathematical symbols you want other than ceiling and floor.

If I remember correctly, the hardest number for me was (I think) 68.

Example:

0 = 99-99.

Good luck!

ACG

P.S.If you want a hint, read on.



















spoiler space.






















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Hint: .9 = 1.

 

[James R]

Hint: .9 = 1.

Err... no it doesn't.

 

[Lyuokdea]

ACG is trying to say that .9 repeating = 1, since you can use the repeating symbol as a mathematical symbol.

~Lyuokdea

 

[Galileo]

9/9+9-9=1

9/9+9/9=2

 

[Jimmy Snyder]

$68 = 69 - \frac{9}{9}$

 

[ACG]

The first few are fairly easy. However, the higher the numbers get, the harder they are. By the time you get to 50-60 you have to start to get VERY creative. Rest assured, it can be done.

 

[ArielGenesis]

jimmysnyder, that's a smart one ^^

 

[neurocomp2003]

but are we allow to rotate the 9? i thoght it woulda changed it into a 6 as it clearly reads six nine not nine nine.

Sqrts allowed?

 

[Learning Curve]

hmmmm... for 2 I thought using .9999 was cheap so i did:


((9*.9)/9)*9

 

[ArielGenesis]

i don't undertand what does .9=1 or .99999 meant

 

[Jimmy Snyder]

i don't undertand what does .9=1 or .99999 meant

I'm not sure but I think it means:

1 = .9...

so

4 = .9... + .9... + .9... + .9...

 

[Daminc]

I think they are referring to .9 (with a dot above the 9 to signify a recurring number)

I don't really agree because it will never equal 1 it just will be very close.

In general terms is it cheating to use the shortened form of a square root (e.g. remove the 2)?

 

[Daminc]

19 down 113 to go :(

 

[ACG]

One may not turn the 9's upside down. Sorry about that!

Square root is legal (and indeed, is crucial). Any mathematical symbol which does not involve letters (sin, cos, etc.) is permitted. Ceiling and floor are prohibited.

The .9=1 thing is

.9 with a bar over it (.9 bar) = .999999... = 1.

ACG

 

[ACG]

Here's another hint -- it took me maybe 3 minutes to figure out the solutions for all of the numbers except maybe five or six. Those six took an hour more. Clearly, there's a trick. But what is it?

 

[neurocomp2003]

3minutes? wow I'm over and hour and I've gotten maybe 3/4s


hour later and I'm down to these #
44,46,65,67,68,74,76,116,
and 116+(i do have a few of these numbers but not alot)


5min later and I'm left with
65,67,78,74,76,116,118,131 i give up for now

 

[Daminc]

The .9=1 thing is

.9 with a bar over it (.9 bar) = .999999... = 1.

Is the bar the same as having a dot above the number?

p.s. as I said before, it will still never=1 it will just be really close.

Why would you say it equals 1?

 

[godchuanz]

Proof of 0.999999999... = 1

Let 0.9999999999... = x

Therefore, we have
10x = 9.99999999999...
10x = 9 + 0.9999999999...
10x = 9 + x
9x = 9
x = 1

x = 0.9999999999999... = 1

 

[Daminc]

Nice bit of trickery
but 0.99999<1 just like 1.000001>1

 

[Jimmy Snyder]

Nice bit of trickery
but 0.99999<1 just like 1.000001>1

what is $1 - .9999... ?$

 

[ACG]

I'm willing to post the solutions on the board if everyone agrees.

In the meantime, I found a second problem.

It's like the Four Nines Problem, except that it's got two 4's and two 9's.

You can get all the numbers from 0 to 232. This is quite a bit harder as there are more options.

 

[Daminc]

what is 1-.9999

Mmmm, good point

It's still 0.0001 (with a recuring dot above the zero) which is >0

I agree it's close enough to zero that it makes no practical difference but I still can't agree that it's zero.

 

[Jimmy Snyder]

It's still 0.0001 (with a recuring dot above the zero)

There is no such mathematical notation. The dot over the zero literally means there is no 1, just 0s. Besides, just as a practical matter, I don't see how you can justify that final 1 in terms of the subtraction problem I set you.

 

[Daminc]

There is no such mathematical notation.

I used to use a recuring dot when I was at school 20 odd years ago.

I never used it in this context but in a way you're dealing with infinities so if you want to have a number that has an infinate number of nines after the decimal then I can have an infinate number of zeros before the 1 :)

Anytime you fix the number of nines then the sum can be calculated.

 

[Jimmy Snyder]

I used to use a recuring dot when I was at school 20 odd years ago.

When I said that there was no such mathematical notation, I was not talking about the dot, I was talking about the 1 at the right hand end after the dot.

I never used it in this context but in a way you're dealing with infinities so if you want to have a number that has an infinate number of nines after the decimal then I can have an infinate number of zeros before the 1 :)

After good, before bad. I'm allowed to put as many 9s as I want after the decimal point because there is a decimal point, but you are not allowed to put any zeros before the last 1 because there is no last 1. That is what infinity means. in = no, finite = end, infinite = no end = no last digit.

Q: What comes after the last zero?
A: There is no last zero.
Q: What comes before the last one?
A: There is no last one.

I still don't see where that 'last' 1 comes from anyway. Certainly not from the subtraction problem I asked you to perform.

 

[Daminc]

1-.9 = 0.1
1-.99 = 0.01
1-.999 = 0.001
1-.9999 = 0.0001

etc

for every additional 9 there would be an additional 0

 

[Jimmy Snyder]

1-.9 = 0.1
1-.99 = 0.01
1-.999 = 0.001
1-.9999 = 0.0001

etc

for every additional 9 there would be an additional 0

But 0.0001 is not the final answer. The final anwer is smaller than 0.0001!

 

[Daminc]

But more than 0 :)

 

[Jimmy Snyder]

0.00001 + .99999 = 1.
What is 0.0000...1 + .9999...?
Looks to me like .9999...1.
What do you get?

 

[Jimmy Snyder]

But more than 0 :)

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

 

[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 :)