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Dec 5th, 2001, 01:40 AM
#1
Thread Starter
Conquistador
4 Numbers, to equal others
OK
i have a bit of a maths problem
I want to find, from 1 to 100 how four 4's can be used to make that number
e.g
1 = (4+4) / (4+4) or (4/4) / (4/4)
2 = (4/4) + (4/4)
and so on
I have worked out all, but the following:
35,
37,
39,
41,
43,
45,
51,
53,
55,
57,
71,
73,
75,
77,
83,
85,
87,
89,
91,
93,
99,
If anyone can help with the solution to any of these, please post
BTW:
Factorial, e.g. 4! = (24) can be used, so can Root(4) = (2), squares, cubes and powers can be used
e.g.
4^2, 4^3, 4^4, 4^x
etc
please help with the above numbers
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Dec 5th, 2001, 05:01 AM
#2
Thread Starter
Conquistador
ahh come on?
no one knows?
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Dec 5th, 2001, 06:35 AM
#3
Thread Starter
Conquistador
ahh come on, where are all you gurus?
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Dec 5th, 2001, 06:54 AM
#4
Addicted Member
35=(4+(4/4)+(4/4))*(4+(4/4)+(4/4)) - (4/4)
or ((4!)/4)*((4!)/4) - (4/4)
does this count?
If it does then the others should be easy - just a bit long
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Dec 5th, 2001, 07:23 AM
#5
Thread Starter
Conquistador
no,
only 4 numbers can be used
e.g.
26 can be
4! + 4 - 4 + Sqr(4)
which is basically 24 + 2 = 26
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Dec 5th, 2001, 07:31 AM
#6
Addicted Member
ah right, sorry I missed that.
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Dec 5th, 2001, 07:54 AM
#7
Thread Starter
Conquistador
ohhh
i thought you had some solutions for me
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Dec 5th, 2001, 07:58 AM
#8
Frenzied Member
35 is (4^3 + 4 + sqr(4)) / sqr(4)
Harry.
"From one thing, know ten thousand things."
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Dec 5th, 2001, 09:14 AM
#9
Hyperactive Member
i remember reading it is possible for all but two numbers in the range 1-99.
can't u just write a program to do it?
There are 10 types of people in the world - those that understand binary, and those that don't.
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Dec 5th, 2001, 09:26 AM
#10
Re: 4 Numbers, to equal others
Originally posted by da_silvy
OK
I have worked out all, but the following:
35,
37,
39,
41,
43,
45,
51,
53,
55,
57,
71,
73,
75,
77,
83,
85,
87,
89,
91,
93,
99,
If anyone can help with the solution to any of these, please post
BTW:
.....and powers can be used
e.g.
4^2, 4^3, 4^4, 4^x
etc
Heres a jumping off point for some:
85 = 4^3 + 4^2 + 4^1 + 4^0
83 = 4^3 + 4^2 + 4^1 - 4^0
77 = 4^3 + 4^2 - 4^1 + 4^0
75 = 4^3 + 4^2 - 4^1 - 4^0
etc...
-Lou
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Dec 5th, 2001, 10:41 AM
#11
Fanatic Member
Is this also allowed?
99 = 4^3.5 - 4^2.5 + 4 - 4^0
You didn't say powers should be integers (besides, sqrt is also used, which is a power of 0.5)
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Dec 5th, 2001, 10:45 AM
#12
Looks like its legal to use non-integer values.
Or, for that matter, (a + sqr(b)), (a - sqr(b)) values. with the proper combination, ALL integer numbers could be generated this way.
{ Although, I personally think only 4's should be used for any number. IMHO }
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Dec 5th, 2001, 01:21 PM
#13
transcendental analytic
hmm, whats the first number that can't be described this way
99=4!*4+4-4^0
93=4!*4-4+4^0
91=4!*4-4-4^0
89=4^3+4!+4^0*4^0
87=4^3+4!-4^0*4^0
73=4^3+4*sqr(4)+4^0
71=4^3+4*sqr(4)-4^0
57=4^3-4-4+4^0
55=4^3-4-4-4^0
53=4^3-4^2+4+4^0
51=4^3-4^2+4-4^0
45=4!+4!-4+4^0
43=4!+4!-4-4^0
41=4!+4^2+4^0*4^0
39=4!+4^2-4^0*4^0
37=24+4^2-sqr(4)-4^0
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