Puzzle: Make 1-50 using only 1,2,3,4 and basic operators

[DavidLeese]

Hello all,
I'd like to share a puzzle, a partial solution, and a request for some more help (or comments).

The puzzle is from an old maths textbook pitched at 12-14 year olds, but has potential for expansion into more difficult territory.

Quote: "Creating Numbers: a task requiring imagination
Your task is to create every number from 1 to 50. You can use only the numbers 1, 2, 3 and 4 once in each and the operations + - x / .

You can use the numbers as powers, and you must use all of the numbers 1, 2, 3, 4. Here are some examples:


1 = (4-3) / (2-1)

20 = 4² +3 +1

68 = 34 x 2 x1

75 = (4+1)² x 3

End Quote.

I've managed the task of generating all the numbers from 1 to 50, and having done that, I've started moving on to 51-60 and possibly further.

My full work so far, and the gaps I've encountered between 51 and 60 are here:
http://davechessgames.blogspot.com/2015/06/numbers-from-1234-to-50.html

Can anybody help me with my gaps? Or - alternatively - prove that the gaps can't be completed?

Thanks

David

 

[DavidLeese]

Thanks, Denis - for the suggestions, comments, and additional solutions for 57.

I have made the correction you highlighted, and changed all the x to * thanks for reminding me on the correct symbol to use online

I'll keep number-crunching and see what else I can get.

Kind regards

David

 

[DavidLeese]

Thanks

Thanks, Denis, for your reply.

I'm not sure why factorial wasn't permitted in the original rules - maybe the textbook authors didn't think their target audience would know about factorials. In any case, I've decided that we need factorials - and decimals - to continue the game!

Thanks for your solutions for 57, and the correction to 28 - I have added these to the blog, and I've also changed the multiplication symbol from x to * thanks for reminding me of correct notation.

I'll keep number-crunching and see if there are any more gaps I can fill in with decimals and factorials!

Kind regards

David

 

[lookagain]

DavidLeese,

here are some expressions for the remaining numbers:


53 = (1 + 4!)2 + 3

53 = (3 + 4!)2 - 1


56 = (1 + 4!)2 + 3!

56 = (1 + 3 + 4!)2

56 = 41 + 3/.2

56 = 4! + 3/.1 + 2

 

[lookagain]

What are your gaps from 61 to 100?
61 to 70 easy enough;
you can get all of them using 4^3, except:

64: 2^(3+4-1) \(\displaystyle \ \ \ \ \ \ \ \ \)not true regarding the "except"
68: 34 *2/1
69: 3^4 - 12\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \)not true regarding the "except"
70: 2 * (34 + 1)\(\displaystyle \ \ \ \ \ \ \)not true regarding the "except"

64: \(\displaystyle \ \) (2 - 1)*4^3

69: \(\displaystyle \ \)4^3 + 1/.2
70:\(\displaystyle \ \) 4^3 + (1 + 2)!

But 99 looks challenging.

99:\(\displaystyle \ \)123 - 4!

 

[Otis]

Your task is to create every number from 1 to 50. You can use only the numbers 1, 2, 3 and 4 once in each and the operations + - x / .

You can use the numbers as powers, and you must use all of the numbers 1, 2, 3, 4. Here are some examples:

68 = 34 x 2 x1

In the example above, the author has violated rules as written (shown in red) because the number 34 is not one of the four allowed numbers AND the numbers 3 and 4 were not used.

Instead, the author ought to have written 'the digits 1,2,3,4'. :cool:

 

[Otis]

Any reason why the factorial ! is not allowed, but the power ^ is?

I've also seen use of the decimal point; like (4 + 2) / .1 - 3 = 57.

Denis regularly tries to change the posted exercise.

 

[Otis]

David, I noticed that you haven't used fractional exponents. I bet you could jazz up some of those which you mentioned (in your blog post) are contrived.

EG:

2 = 4^(3/2 - 1)

9 = 4^(3/2) + 1

If not, here's another thought:

35 = 2^(4 + 1) + 3

Cheers :cool:

 

[DavidLeese]

Many thanks to everyone who has replied to this thread - I wasn't sure what sort of reception I'd get, so the warm welcome is very kind.

I have changed the phrasing of the question - it now talks about 'digits' instead of 'numbers'
I have also added all of the solutions between 51 and 100, and come up with some more of my own.

I'm still working on 'jazzy' solutions, my current favourite is 90 = (1 + 2) * (3! + 4!) since it gets two factorials into the solution.

At this point, I'm missing only 94, 97, 98 and 100. There are bonus marks for answers that are perceived to be elegant (it's all completely arbitrary, and we can argue about the nicer or jazzier answers separately) - for instance, I was able to get most of the answers without having to use decimals ;-)

Link: http://davechessgames.blogspot.co.uk/2015/06/numbers-from-1234-to-50.html

Thanks again

David

 

[lookagain]

I'm still working on 'jazzy' solutions, my current favourite is 90 = (1 + 2) * (3! + 4!) since it gets two factorials into the solution.

DavidLeese, this gets four factorials into the solution:


90 = (1! + 2!)(3! + 4!)

At this point, I'm missing only 94, 97, 98 and 100.

100 easy enough: 3/.12 * 4 = 100.

There was already an expression for 100 offered above, unless you insist it be without a decimal point (if it exists in that form).


- - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - -- - - - -- - - - - - - - - - - - - - - - - - - - - -- - -- - - - - -- - - -


Anyway, here are solutions for 94, 97, 98, and 100 that do not use decimal points:


94 = (1 + 3)4! - 2

94 = 4(1 + 3)! - 2


97 = 4(3! - 2)! + 1


98 = (1 + 3)4! + 2

98 = 4(1 + 3)! + 2


100 = (1 + 4!)(3! - 2)

 

[lookagain]

Here's a LEGAL solution (plus introduction of double factorial):
89 = (3!)! / (2 * 4) - 1

That is not a "double factorial." That has a special look, and it is a
special function unto itself.

You can read here: http://mathworld.wolfram.com/DoubleFactorial.html


You made use of a repeated application of the factorial.


-- - - -- - - - - -- - - - - -- -- - - - - - -- - - - - - - - - -- -


-----> DavidLeese, not only for 89, but you also have an invalid solution for 95. <-----


You must have had "46" on your mind.


- --- - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - -- - -


Here's an alternate expression for 89, where the digits are in ascending order:


89 = 1 + 2^(3!) + 4!




I found these for 95:


95 = 3!*(2^4) - 1

95 = 3!*(4^2) - 1

 

[lookagain]

So?
(3!)! = 720 ; who cares about nomenclature...
http://www.wolframalpha.com/input/?i=(3!)!=

Your question makes zero sense. You called it the wrong thing, and I corrected you.
Live and learn.

And be grateful you learned something new about the terminology.


And by the way, if I had not cared, I wouldn't have bothered to post it.

1) Don't get defensive.

2) Thank the messenger for it and apply it if it comes up again.

 

[greg1313]

89 = (3!)! / (2 * 4) - 1

Good solution!

 

[Steven G]

Wow! To give credit where it's due:
you are the Wayne Gretzky of this puzzle type!

WARNING: if you ever play in a hockey game,
make sure I'm not on the other team !

I am a combination of Giacomin and Villemure so bring it on.

 

[DavidLeese]

Good work David!

Noticed you have 89 = 46 * 2 - 3 ... you're cheating!
Here's a LEGAL solution
89 = (3!)! / (2 * 4) - 1

Errr... yeah... I'm not exactly cheating, I'm just reworking the rules for my own personal benefit Thanks for catching my error, and for the alternative solution, which is probably the mathematical equivalent of a double-letter score and a triple word score.

You could have a simpler version for 71-73:
71 = 24 * 3 - 1
72 = 24 * 3 * 1
73 = 24 * 3 + 1

I tried to steer clear of these answers... I thought it was a little lazy to constantly have sequences of ...
Answer = X - 1
Answer2 = X * 1
Answer 3 = X +1

... so if I could find a different type of solution, then I'd use that instead. Having said that, I like your solution to 72 because it's a simple product, so I'm going to nab that one :-D

And for 87:
87 = 43 * 2 + 1

And for 78:
78 = 312 / 4[/QUOTE]

Thanks, Denis!

 

[DavidLeese]

Anyway, here are solutions for 94, 97, 98, and 100 that do not use decimal points:


94 = (1 + 3)4! - 2

94 = 4(1 + 3)! - 2


97 = 4(3! - 2)! + 1


98 = (1 + 3)4! + 2

98 = 4(1 + 3)! + 2


100 = (1 + 4!)(3! - 2)

Great, thanks very much!

I feel like I (or we) have completed the set... does anybody want to do 100+? I'll see how many I can get from 100 to 150 over the next few days, and if I think I'm making progress, I'll share an update. Otherwise, I think I'll revisit the maths textbook and see what else I can find to get the brain cells working.

 

[DavidLeese]

101- to 150

Hello again, mathematicians.

This has taken slightly longer than anticipated, but I'd like to share an update on the 1-100 with 1,2,3,4 puzzle that we discussed last month. I've made some progress on 101-150, finding solutions for 37 of the 50 numbers.

My solutions, and the unsolved numbers, are here:
http://davechessgames.blogspot.com/2015/07/numbers-1234-from-100-to-150.html

Please can you help me fill the gaps? In particular, the odd numbers between 100 and 114 are proving very difficult (no easy way of multiplying up to them).

Also, as always, if you find an error anywhere, please let me know.

Thanks

David

 

[DavidLeese]

Thanks, Denis.

Since we're going way beyond the original scope of the question, I figure we can use any sensible functions in our search for the solutions.

Am I right in thinking that CEILING is the equisvalent of the Excel function ROUNDUP, and FLOOR is the equivalent of ROUNDDOWN (the modern equivalent of the BASIC function INT) ?

I'll include your many solutions into my list, and cite you as the solver.

Thanks

David

 

[lookagain]

Thanks, Denis.

Since we're going way beyond the original scope of the question, I figure we can use any sensible functions in our search for the solutions.

Am I right in thinking that CEILING is the equisvalent of the Excel function ROUNDUP, and FLOOR is the equivalent of ROUNDDOWN
(the modern equivalent of the BASIC function INT) ?

I'll include your many solutions into my list, and cite you as the solver.

Thanks

David

DavidLeese, please use my versions instead for your list:
-----------------------------------------------------------------------


103: \(\displaystyle \ \ \dfrac{\sqrt{4}}{.1*.2} \ + \ 3\)

105: \(\displaystyle \ \ \dfrac{4! - 3}{2(.1)}\)

107: \(\displaystyle \ \ \dfrac{4!}{.2} - 13\)

109: \(\displaystyle \ \ \dfrac{4! - 1}{.2} - 3!\)

111: \(\displaystyle \ \ (1 + 4)! - 3^2\)

112: \(\displaystyle \ \ (1 + 3!)*4^2\)

113: \(\displaystyle \ \ \dfrac{4!}{.2} - 1 - 3!\)

131: \(\displaystyle \ \ \dfrac{1 + 4!}{.2} + 3!\)

137: \(\displaystyle \ \ \ no \ \ new \ \ form \ \ for \ \ it \ \ (Refer \ \ back \ \ to \ \ Denis's \ \ solution.)\)

140: \(\displaystyle \ \ \dfrac{1 + 3 + 4!}{.2}\)

149: \(\displaystyle \ \ \dfrac{4! + 3!}{.2} - 1\)

 

[DavidLeese]

Outstanding! Thanks so much for your solutions. I have [finally] updated my blog with the solutions you have provided.

From what I can see, only 133 remains without any solution, which is an amazing result considering how far we are beyond the original scope of the first question (solve for 1-50).

Any takers for 133?

Thanks

David