check this out...
by Try this out:
Use four 4's and add, subtract, multiply, divide and square root them to
try to come up with every number from 1 to 100. You must use all four
4's, and no more than four. You can not square any of the fours to make
it 16, but you can raise a four to the fourth power.
You can also use factorials. Like (4! * 4 + 4) / 4
4! is equal to 4 * 3 * 2 * 1 which equals 24.
and you can use 4 to the 4th power. <-- This counts as using 2 fours.
You put the square root sign over any part of the equation. It is
hard to illustrate on the computer, but I hope you can see what I'm doing
here:
(square root of (4) * 4!) / (square root of (2)) + 4! = 72
This is how you get the number 1:
(4 + 4 - 4) / 4 = 1.
Now try to figure out how to get the number 2. Then 3. Up to 100.
It is possible. Some are very easy, some are medium, and some are very
difficult to figure out.
...
You can also group the fours into the numbers 44, 444 and 4444. But if you
use 4444, then you've used up all your fours.
If you can get 50 or more, E-Mail them to me.
If you get all 100, you are awesome.
AI Summary
13 Comments
my hair hurts just reading that!
yeah, this is from some old brain-teaser that ian responded to. i remember him showing us a google search on his name and his response to this was in a Usenet group.
haha... man! you blew my cover! yeah, guys this is ian's from like 1996
actually, it was an extra credit problem in my math class, and i couldn't get all the answers so i asked in a math newsgroup
ahhh. so you wanted extra credit so you tried to fool everyone in the newsgroup into helping you out by telling them it was a neat-0 brain-teaser. slick.
Matt your a dork dude
yeah i always knew he was a dork, back when he brought out htat "Mole D. Kelly" article.
if i was awake enough to read this, i STILL wouldnt even try to do it... blah blah blah, matt kelly is half man half beanstalk
Since the bases and operators are finite, this puzzle can be solved by a
brute force method. According to the question, the basis is 4. With unitary
operations, i.e., n!, and SQR(n), we now have v1= < 4, 24, 2 > as our basis.
By using binary operations, i.e., +, -, *, /,and ^ we can generate
the following matrix:+ 4 24 2
4 8 28 6
24 +/-20 48 26
2 +/- 2 +/-22 4
0
Similarly,* 4 24 2
4 16 96 8
24 6 576 48
2 2 12 4
1And so on...
So by using two 4's, we have v2=<-22, -20, -2, 0, 1/12, 1/6, 1/2, 1
2 4 6 8 12 16 20 22 26 24 26 28 (44) 48 96 256, ... 24^24> as our basis.
(After sorting, and ignoring SQR(v2) and (v2)! for simplicity.)
By using v1 and v2, we can proceed to generate v3 with the same
method. v3 should consist all the possible numbers that can generate
from three 4's with the operators mentioned. We then can generate all
the possible number that can form from four 4's and the operators by
v4=SORT(operator(v3,v1)/operator(v2,v2)). (/=logical AND)
(ignoring (v4)!, SQR(v4), ((v4)!)!, SQR(SQR(v4)), ... etc)To save time, I found SORT(+(v2,v2)/-(v2,v2)/*(v2,v2)) will generate
all the number between 0-> 100 except:
11 15 19 23 31 33 35 37 39 41 43 51 53 55 57 58
59 61 62 63 65 67 69 71 73 75 77 78 79 81 82 83
84 85 86 87 89 90 91 93 95 and 99. (That's more than 50 generated.)
62 and 90 belong to operator(v3,v1):
4*4*4-2=62 and (4!/4)/4/SQR(4)=90 and other numbers in the 'miss-list'
is also belong to operator(v3,v1). Those numbers that do not appear
in v4 and unitary operations on v4 can NEVER be generate by four 4's
and the operators. Mr. Kelly don't think anyone is awesome. ;)
you are a dork
now my eyes are bleeding
i want my mommy.
haha, i had to do this in my math class. For everyone you got right you got like extra credit points.