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So call 3^3 = a and 2^2^2 = b. Now,
call 3^a = a1 and 2^b = b1, now we can continue
Call 3^a1 = a2 and 2^b1 = b2. We can continue recursively like this until we work our way down the tower. So
Welcome to the forum!
]]>Seems like one day you'll be able to work upon them blindfolded!
3. Which is greater?
please give a solution to this problem.
]]>Welcome to the forum. Please post your math problems in the Help me section. You will get all the help you want.
]]>Finding n's maximum value is neat! Good logic...I'd never have thought of that. My method was to record the length of the highest-value number by comparing the results during the calcs.
Re finding all Ns...from your info I can't see how the initial 3/6 and 4/5 numbers are established before going cycling with them, but unless there's an astounding logical solution I guess I don't really have to know.
Nice problem, btw!
]]>I'd been hoping I'd got it right.
By all means! When I asked my colleagues this questions and found a mismatch, I told them that the correct answer has to be for the reasons you can understand!
The method they used was....
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Thanks for that...I'd been hoping I'd got it right.
Do you know how your workmates came up with their results? Is there a good logical method I missed?
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Have you had a chance yet to verify my answer to the first part of Q4? Just curious as to whether or not I got it right. Ta.
]]>...for the second part perhaps "n" is standing upon its head?
Oops! Tricky blighter...doing handstands while I wasn't looking! I've stood him back up on his feet again!
]]>4. "How many such N's exist?"
EDIT: a different output of the same solution. I saw a pattern in the original output and altered the program (BASIC), which now completes in a tiny fraction of the original time.
"What is the maximum value of 'n'?"
]]>You might have considered it as "1-digit N, 2-digit N, 7-digit N etc." where all "n"s are different. Instead, consider it as "4-digit N: 1234, 3456, 4567" where "all the 4-digits" of N are distinct.
For 3, its a Power Tower. 2 is raised by 1001 two's and 3 is raised by 1000 three's. You don't start counting from the "base" of the exponent in counting 1001/1000.
Thus, 2² has 2 raised to 2, 1 times.
Is the explanation good?
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