I am not getting them either!
]]>2667 7662
2710 8757
I have employed all the familiar operators, plus, minus, divide, mutiply, roots, exponents, factorials and have for about the last twenty or so gone to the use of DOUBLE factorials (like 7675 being 7! / 6! = 7!! / 5!!), and in a couple of instances DECIMAL notation (like 5657 being .5 = 6 / (5+7)).
If someone could set those last four to bed, I could move on with my life. Thank you!!
]]>When I posted post #11. I did not see your solution else I would not have posted.
What is holding you up about these, bobbym!?
I like this, because it actually solves all the rest as well.
Darby is satisfied, what is there left for me to do?
]]>Hi Darby;
I am sorry but I have been unable to find anything for those
that does not use something more complicated then what
you have been using.
10:38
1 = [0(3 + 8)]!
- - - - - - - - - -- -
10:47
1 = [0(4 + 7)]!
- - - - - - - - - - -
10:57
1 = [0(5 + 7)]!
- - - - - - - - - - -
10:58
1 = [0(5 + 8)]!
- - - - - - - - - -
10:59
1 = [0(5 + 9)]!
__________________
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
]]>Darby wrote:1 = (0 X 2 X 7)!
1 = (0/2 X 7)!
1 = [0 X (2 + 7)]!
1 = [0 X (2 - 7)]!
1 = [0 X 2^7]!
1 = [0/(2^7)]!
1 = (0^27)!
1 = [0^ (2 X 7)]!
1 = [0^(2^7)]!
1 = [(0^2)^7]!
1 = [0^2 X 7]!
1 = [0^(2/7)]!
1 = [(0/2)/7]!
1 = [(0 X 2)/7]!
1 = (0/2/7)!
1 = (0/2 X 7)!
1 = (0/27)!
1 = [0/(2 X 7)]!
1 = [(0 X 2)^7]!
1 = [(0/2)^7]!
1 = [0/(2^7)]!
1 = [0/(2 + 7)]!
I like this, because it actually solves all the rest as well. I went over all my equations to see if I had used zero as an exponent (another defined term) and I hadn't - and these fit the criteria so closely they work. Plus....I lack the mathematical depth to continue searching for some elusive root of some elusive factorial....
Thanks!!
]]>I am sorry but I have been unable to find anything for those that does not use something more complicated then what you have been using.
]]>1 = (0 X 2 X 7)!
1 = (0/2 X 7)!
1 = [0 X (2 + 7)]!
1 = [0 X (2 - 7)]!
1 = [0 X 2^7]!
1 = [0/(2^7)]!
1 = (0^27)!
1 = [0^ (2 X 7)]!
1 = [0^(2^7)]!
1 = [(0^2)^7]!
1 = [0^2 X 7]!
1 = [0^(2/7)]!
1 = [(0/2)/7]!
1 = [(0 X 2)/7]!
1 = (0/2/7)!
1 = (0/2 X 7)!
1 = (0/27)!
1 = [0/(2 X 7)]!
1 = [(0 X 2)^7]!
1 = [(0/2)^7]!
1 = [0/(2^7)]!
1 = [0/(2 + 7)]!
]]>10:27, 10:38, 10:47, 10:57, 10:58, 10:59
If you can get those, you are better than I am (like that hasn't been proved already....)
Thanks, and have fun.
]]>Hi;
That is the floor function there is also a ceiling function.
I see what you mean. That would be stretching things just a bit, but in the absence of any other exact solutions I suppose it would be OK. As long as everything isn't quickly solved by this application (sort of like my "cheater" way on the 11 o'clock equations where almost every one of them can be written as 1 being equal to a power of one.....)
]]>That is the floor function there is also a ceiling function.
]]>I'm a C+ Algebra 11 student (although Algebra 11 was decades ago....)
]]>10:26
(1+0+2)! = 6
10:28
10 = 2+8
]]>