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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.

]]>There's no logical method known to me to find those 3, 4, 5 and 6 digit Ns.

Your computer program is certainly better and faster too!]]>

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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Sorry for the delay! I did not forget bout it but just was a little bit messed up on somethings. I gave the problem to two of my workmates and they came up with as the total number of numbers possible. The latter one had missed on a few numbers!!

Your solution is flawless!

You Rock!]]>

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!

]]>I'll need to verify the answer to first part and for the second part perhaps "n" is standing upon its head?]]>

4. "How many such N's exist?"

*EDIT*:

"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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