Classic problem: Four 4's - Page 9

Zeta-Flux · 2005-10-03, 00:01

(edited for a couple mistakes - hope to replace soon)

Zeta-Flux · 2005-10-03, 00:52

In 2/4 time:

25 = 4/((.4)^2)

(of course, this means that there are lots of other ones that now are in 3/4 time, like 21=25-4, 29=25+4)

Zeta-Flux · 2005-10-03, 03:51

Oops. Just noticed Xilman's post tucked away there.

Zeta-Flux · 2005-10-03, 04:07

Quote:

Originally Posted by rogue

I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.

93=17^2 - 14^2.

cheesehead · 2005-10-03, 06:36

I'm trying to use 93 = 47²-46², by first expanding the 4-4s representations of 47² and 46², then simplifying by collecting common terms. The expansion out to 39 fours worked just fine. So far I've simplified that down to only 32 fours:

93 = 4!*4!*4 - 4!*4 - 4!*4!/.4^sq + 4!*4/.4^sq + 4!*4/.4^sq - 4^sq/.4^sq + 4!*4/.4 + 4!*4/.4 - 4*4/.4 - 4*4/.4 - 4*4 + 4/4

I suspect that " + 4/4" term isn't going away, so I just have to reduce the other 30 fours down to a two-4s representation of 92.

cheesehead · 2005-10-03, 07:10

Quote:

Originally Posted by cheesehead

93 = 4!*4!*4 - 4!*4 - 4!*4!/.4^sq + 4!*4/.4^sq + 4!*4/.4^sq - 4^sq/.4^sq + 4!*4/.4 + 4!*4/.4 - 4*4/.4 - 4*4/.4 - 4*4 + 4/4

Well, soon after I posted that I realized I could take out six 4s by combining the three terms just before "+ 4/4" to make another "- 4!*4" term:

4!*4!*4 - 4!*4 - 4!*4 - 4!*4!/.4^sq + 4!*4/.4^sq + 4!*4/.4^sq - 4^sq/.4^sq + 4!*4/.4 + 4!*4/.4 + 4/4

... and now we can collect all the terms involving 4! like so:

4!*(4!*4 - 4 - 4 - 4!/.4^sq + 4/.4^sq + 4/.4^sq + 4/.4 + 4/.4) - 4^sq/.4^sq + 4/4

= 4!*(4+4) - 4^sq/.4^sq + 4/4

but that may not have actually helped. :-}

Wacky · 2005-10-03, 11:51

I think that we can show that this is an impossible approach.

If 93 = a² - b²
then (a+b) | 93 and (a-b) | 93 .
Since 93 = 3 * 31, this leads to only two possible solutions,
namely 93 = 17² - 14²
and 93 = 47² - 46²
In both cases, the odd number cannot be generated by less than three fours. (No odd number can be generated in less than two).
But the even number cannot be generated by a single four.
Therefore, there is no solution of this form using at most four fours.

By similar logic, I beleive that we will also eliminate all other possible forms that conform to the set of "rules" that Ken chose for this thread.

cheesehead · 2005-10-03, 15:27

Quote:

Originally Posted by Wacky

In both cases, the odd number cannot be generated by less than three fours. (No odd number can be generated in less than two).

But the even number cannot be generated by a single four.
Therefore, there is no solution of this form using at most four fours.

Counterexample:

3 = 4-4/4

12 = 4^sq-4

15 = 4^sq-4/4

If a term in the fours representation of the odd number can be combined with a term in the fours representation of the even number to either reduce the number of fours necessary in the sum/difference or even cancel each other, then the representation of the sum/difference can have fewer fours than the total number of fours in the two separate representations.

Quote:

By similar logic, I beleive that we will also eliminate all other possible forms that conform to the set of "rules" that Ken chose for this thread.

But we'll have to take the possibilities of simplification and cancellation into account.

cheesehead · 2005-10-03, 15:54

Oh, wait -- you meant that in the particular cases of 17² and 14² or 47² and 46², the odd numbers require at least three and the even numbers require at least two. So 3 + 12 = 15 is not a valid counterexample.

I still don't see why it is not possible that cancellation or simplification could result in a four fours representation of 93 under the current set of rules.

Wacky · 2005-10-03, 16:17

Quote:

Originally Posted by cheesehead

Counterexample:
3 = 4-4/4
12 = 4^sq-4
15 = 4^sq-4/4

This is NOT a case of the form A^sq -B^sq
which is the only case that I addressed.

In your example, you have:
(B-C) + (A-B) which clearly reduces to (A-C)

To avoid the possibility of such a cancellation, we can define a complexity function that rates (A-C) "simpler" than (B-C) + (A-B) and thus, by induction on this complexity, the latter can be a solution only if the former is.

Plese note that I did not claim to have presented a proof for all cases, but I think that it can be developed.

Wacky · 2005-10-03, 16:32

Quote:

Originally Posted by cheesehead

Oh, wait -- you meant that in the particular cases of 17² and 14² or 47² and 46², the odd numbers require at least three and the even numbers require at least two. So 3 + 12 = 15 is not a valid counterexample.

Correct, I have only eliminated the particular case of the form A² - B²

Quote:

I still don't see why it is not possible that cancellation or simplification could result in a four fours representation of 93 under the current set of rules.

It is possible that it could. I suspect that the most difficult form to eliminate will be A!-B².

2005-10-03, 00:01	#89
Zeta-Flux May 2003 3013₈ Posts	(edited for a couple mistakes - hope to replace soon) Last fiddled with by Zeta-Flux on 2005-10-03 at 00:04

2005-10-03, 00:52	#90
Zeta-Flux May 2003 7×13×17 Posts	In 2/4 time: 25 = 4/((.4)^2) (of course, this means that there are lots of other ones that now are in 3/4 time, like 21=25-4, 29=25+4) Last fiddled with by Zeta-Flux on 2005-10-03 at 01:04

2005-10-03, 06:36	#93
cheesehead "Richard B. Woods" Aug 2002 Wisconsin USA 2²×3×641 Posts	I'm trying to use 93 = 47²-46², by first expanding the 4-4s representations of 47² and 46², then simplifying by collecting common terms. The expansion out to 39 fours worked just fine. So far I've simplified that down to only 32 fours: 93 = 4!4!4 - 4!4 - 4!4!/.4^sq + 4!4/.4^sq + 4!4/.4^sq - 4^sq/.4^sq + 4!4/.4 + 4!4/.4 - 44/.4 - 44/.4 - 44 + 4/4 I suspect that " + 4/4" term isn't going away, so I just have to reduce the other 30 fours down to a two-4s representation of 92. Last fiddled with by cheesehead on 2005-10-03 at 06:49*

2005-10-03, 11:51	#95
Wacky Jun 2003 The Texas Hill Country 441₁₆ Posts	93 = a^2 - b^2 I think that we can show that this is an impossible approach. If 93 = a² - b² then (a+b) \| 93 and (a-b) \| 93 . Since 93 = 3 * 31, this leads to only two possible solutions, namely 93 = 17² - 14² and 93 = 47² - 46² In both cases, the odd number cannot be generated by less than three fours. (No odd number can be generated in less than two). But the even number cannot be generated by a single four. Therefore, there is no solution of this form using at most four fours. By similar logic, I beleive that we will also eliminate all other possible forms that conform to the set of "rules" that Ken chose for this thread.

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2005-10-03, 03:51	#91
Zeta-Flux May 2003 7·13·17 Posts	Oops. Just noticed Xilman's post tucked away there.

2005-10-03, 15:54	#97
cheesehead "Richard B. Woods" Aug 2002 Wisconsin USA 2²×3×641 Posts	Oh, wait -- you meant that in the particular cases of 17² and 14² or 47² and 46², the odd numbers require at least three and the even numbers require at least two. So 3 + 12 = 15 is not a valid counterexample. I still don't see why it is not possible that cancellation or simplification could result in a four fours representation of 93 under the current set of rules.