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        riddles >> easy >> Simple equations: 4 digits and a target
        
(Message started by: WombatDeath on May 21^st, 2003, 1:12pm)

Title: Simple equations: 4 digits and a target
Post by WombatDeath on May 21^st, 2003, 1:12pm

Apologies in advance if this has been posted before.

You're given four digits and a target. Your mission is to form half of an equation with these four digits, using only the four basic operators (plus, minus, divide and multiply) that resolves to the given target. You must use each digit exactly once.

An easy example:

Using the digits 2, 3, 5 and 6 exactly once each, make 18.
Answer: ((6 * 5) / 2) + 3 = 18

Now the real questions:

1) Using the digits 1, 3, 4 and 6 exactly once each, make 24.

2) Make 24 again, this time using 3, 3, 7 and 7 exactly once each.

(Note: there's no trick to either of these questions - you don't have to turn the digits upside-down, or convert them to roman numerals, or anything similarly silly)

Title: Re: Simple equations: 4 digits and a target
Post by Kitty on May 21^st, 2003, 1:46pm

No 2 has been posted, here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1052182452)

;D

Title: Re: Simple equations: 4 digits and a target
Post by WombatDeath on May 21^st, 2003, 1:57pm

Bah! And I managed to miss the damn thing right at the top of page two :'(

Ah well, I still haven't found the first one posted here (it's probably easier than the second - still, it's quite a nice little puzzle).

If anyone really has too much time on their hands, there's a similar problem involving four 4s and every integer between 1 and 100 ;>

Title: Re: Simple equations: 4 digits and a target
Post by Icarus on May 21^st, 2003, 3:48pm

WombatDeath - try the search function on the toolbar at the top of the page. It can be handy in finding old puzzles.

Title: Re: Simple equations: 4 digits and a target
Post by Lightboxes on May 29^th, 2003, 4:37pm

Using 1,3,4,6 to make 24 is impossible to me...
Reasons:
[hide]The number 1 actually isn't needed or even helpful. 1 will leave the answer the same if it x or / if you + or - then the 2nd to last step would have to be used on a 23 or 25. 23 and 25 are unattainable with just 3,4,6 because the only way to get 25 is to x 6 with a number higher than 4. 3 and 4 cannot reach 25. 25 cannot be reached by 3x4 and then using the 6 to get 25.
In other words...3,4,6 need to be multiplied eventually to get 23 or 25, which wouldn't work. AND 1 won't even help, in any situation, unless you use it to make a number like 2.3333 (4/3 + 1) but then you've used up all the numbers and can't make the 6 jump to 24 by x.[/hide]

Title: Re: Simple equations: 4 digits and a target
Post by MattyDK23 on May 29^th, 2003, 6:37pm

The answer is:[hide]
6 / (1 - (3 / 4)) = 6 / 0.25 = 24

That 1 was the key...and when I realized that most of my attempts were using multiplication or addition for the 'final' operation (and failing miserably), I changed it to division -- realizing I needed a number less than 1 as the denominator.
[/hide]

Title: Re: Simple equations: 4 digits and a target
Post by Lightboxes on May 29^th, 2003, 7:25pm

stupid me...I got so obsessed with increasing the number by multiplication...I forgot to use a number and divide it by a decimal.

Title: Re: Simple equations: 4 digits and a target
Post by WombatDeath on May 30^th, 2003, 1:09pm

Well done Matty - it's one of those which is blindingly obvious once you know the answer, but most people (including myself) seem not to consider division as the important step. If anyone wants a longer puzzle along similar lines you might try this:

Using the number 4 four times, make every integer between 1 and 100. This time you have more operators available to you. You can use the following as often as you like:

+ (addition)
- (subtraction)
* (multiplication)
/ (division)
sqrt (square root)
** (exponent)
! (factorial)

And as well as that, you can also use 4/10 (.4) and 4/9 (.4444444...). Note that you must use arithmetic, and not concatenation - you can't stick two 4s together to get 44.

Example: to get 75 you could do (4! + 4 + (sqrt(4)))/.4

Title: Re: Simple equations: 4 digits and a target
Post by Sir Col on May 31^st, 2003, 7:22am

Teehee, this is a favourite for my classes. It really sorts the wheat from the chaff!

However, rather than re-invent the wheel, you might like to check-out this (http://www.comp-sci.demon.co.uk/FourFours_So.html).

I think more interesting questions for this puzzle are:
(i) What is the first number that cannot be derived?
(ii) What is the largest prime you can make?

Question (ii) is interesting as you could use nested factorials to create VERY large numbers, e.g. ((4!)!)!. The best I've found so far is (4*4-4)!–[sqr(sqr(4))] = 479001599, but I am using the integer part function.

BTW, using logs I can show how it is possible to obtain any finite value at all.

Title: Re: Simple equations: 4 digits and a target
Post by Leonid Broukhis on May 31^st, 2003, 9:15am

Using logs, you really only need three 2's to express any finite non-negative number.

Title: Re: Simple equations: 4 digits and a target
Post by Sir Col on May 31^st, 2003, 2:08pm

... but the challenge is to do it with exactly four fours – no more and no less.

Title: Re: Simple equations: 4 digits and a target
Post by Leonid Broukhis on Jun 1^st, 2003, 1:57am

on 05/31/03 at 14:08:57, Sir Col wrote:

... but the challenge is to do it with exactly four fours – no more and no less.

...which is trivial after one figures out how to do it with three twos.

Title: Re: Simple equations: 4 digits and a target
Post by Sir Col on Jun 1^st, 2003, 2:13am

*feels stupid and pointing to your avatar* "Go on, say it to me, Homer."

Title: Re: Simple equations: 4 digits and a target
Post by Leonid Broukhis on Jun 1^st, 2003, 8:02am

Without revealing too much:

Three fours are under square roots to represent the three twos I was talking about, and are used to express the number that is 4 more than desired; the last 4 is subtracted from the expression.

Title: Re: Simple equations: 4 digits and a target
Post by Sir Col on Jun 1^st, 2003, 9:15am

My method works as follows:

sqr(4)=4^1/2, sqr(sqr(4))=4^1/4, sqr(sqr(sqr(4)))=4^1/8

Therefore log_4(sqr(4))=1/2, log_4(sqr(sqr(4))=1/4, log_4(sqr(sqr(sqr(4)))=1/8, et cetera.

log_2(x) can be obtained by using 4/sqr(4)=2.

Hence log_2(log_4(sqr(4)))=1, log_2(log_4(sqr(sqr(4))))=2, log_2(log_4(sqr(sqr(sqr(4)))))=3, and so on.