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Topic: Simple equations: 4 digits and a target (Read 795 times) |
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WombatDeath
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Simple equations: 4 digits and a target
« on: May 21st, 2003, 1:12pm » |
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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)
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Kitty
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Re: Simple equations: 4 digits and a target
« Reply #1 on: May 21st, 2003, 1:46pm » |
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No 2 has been posted, here
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If the human brain was simple enough for us to understand, we would still be so stupid that we couldn't understand it.~Kant (1724-1804)
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WombatDeath
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Re: Simple equations: 4 digits and a target
« Reply #2 on: May 21st, 2003, 1:57pm » |
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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 ;>
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| « Last Edit: May 21st, 2003, 1:58pm by WombatDeath » |
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Icarus
wu::riddles Moderator
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Re: Simple equations: 4 digits and a target
« Reply #3 on: May 21st, 2003, 3:48pm » |
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WombatDeath - try the search function on the toolbar at the top of the page. It can be handy in finding old puzzles.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Lightboxes
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Re: Simple equations: 4 digits and a target
« Reply #4 on: May 29th, 2003, 4:37pm » |
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Using 1,3,4,6 to make 24 is impossible to me...
Reasons:
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.
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| « Last Edit: May 29th, 2003, 4:39pm by Lightboxes » |
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MattyDK23
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Re: Simple equations: 4 digits and a target
« Reply #5 on: May 29th, 2003, 6:37pm » |
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The answer is:
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.
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Lightboxes
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Re: Simple equations: 4 digits and a target
« Reply #6 on: May 29th, 2003, 7:25pm » |
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stupid me...I got so obsessed with increasing the number by multiplication...I forgot to use a number and divide it by a decimal.
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WombatDeath
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Re: Simple equations: 4 digits and a target
« Reply #7 on: May 30th, 2003, 1:09pm » |
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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
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Sir Col
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Re: Simple equations: 4 digits and a target
« Reply #8 on: May 31st, 2003, 7:22am » |
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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.
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.
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| « Last Edit: May 31st, 2003, 7:24am by Sir Col » |
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Leo Broukhis
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Re: Simple equations: 4 digits and a target
« Reply #9 on: May 31st, 2003, 9:15am » |
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Using logs, you really only need three 2's to express any finite non-negative number.
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Sir Col
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Re: Simple equations: 4 digits and a target
« Reply #10 on: May 31st, 2003, 2:08pm » |
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... but the challenge is to do it with exactly four fours – no more and no less.
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Leo Broukhis
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Re: Simple equations: 4 digits and a target
« Reply #11 on: Jun 1st, 2003, 1:57am » |
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on May 31st, 2003, 2:08pm, Sir Col wrote:| ... but the challenge is to do it with exactly four fours – no more and no less. |
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...which is trivial after one figures out how to do it with three twos.
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Sir Col
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Re: Simple equations: 4 digits and a target
« Reply #12 on: Jun 1st, 2003, 2:13am » |
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*feels stupid and pointing to your avatar* "Go on, say it to me, Homer."
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Leo Broukhis
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Re: Simple equations: 4 digits and a target
« Reply #13 on: Jun 1st, 2003, 8:02am » |
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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.
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Sir Col
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Re: Simple equations: 4 digits and a target
« Reply #14 on: Jun 1st, 2003, 9:15am » |
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My method works as follows:
sqr(4)=41/2, sqr(sqr(4))=41/4, sqr(sqr(sqr(4)))=41/8
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.
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| « Last Edit: Jun 1st, 2003, 9:16am by Sir Col » |
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