Problem with quiz

[kemik]

How to solve this quiz? http://mathzone.net.tc
Please help, it is my homework...

[ewmayer]

The whole point of homework is for the student to solve problems using their own skill, creativity and formal learning, so I'll restrict myself to a few broad hints, especially ones with a number-theoretic flavor, in keeping with the tenor of this forum.

I was able to express 2005 using 17 zeros (one more than the best-claimed solution) and standard arithmetic operators in several different ways. You might try looking at the factorization of 2005 (as well as integers near 2005), expressing 2005 as the sum or difference of small powers, and so forth. In doing so it's useful to build a small table counting the minimum number of zeros needed to express various small integers according to the problem rules, e.g. 1 = 0! needs just one zero, 8 = 2^3 needs 5, and so forth.

Have fun,
-E

[xilman]

Originally Posted by ewmayer

The whole point of homework is for the student to solve problems using their own skill, creativity and formal learning, so I'll restrict myself to a few broad hints, especially ones with a number-theoretic flavor, in keeping with the tenor of this forum.

I was able to express 2005 using 17 zeros (one more than the best-claimed solution) and standard arithmetic operators in several different ways. You might try looking at the factorization of 2005 (as well as integers near 2005), expressing 2005 as the sum or difference of small powers, and so forth. In doing so it's useful to build a small table counting the minimum number of zeros needed to express various small integers according to the problem rules, e.g. 1 = 0! needs just one zero, 8 = 2^3 needs 5, and so forth.

Have fun,
-E

I could manage 2005 with just four 4s and two 2s

Paul

[Uncwilly]

Originally Posted by xilman

I could manage 2005 with just four 4s and two 2s

I cannot conceive how two 2's with any combination of operators can yield 2005.
Paul, the way that I understand it is to use only a selected digit (as many times as you wish), but only that digit.


Kemik: This thread: http://www.mersenneforum.org/showthread.php?t=2007 might be of help.
Also, notice that x/x = 1, which is often handy in these kinds of things.

[akruppa]

I managed 2005 with six 4s - one better than their current best.

The whole thing is kinda pointless if they don't restrict the functions you can choose from - obviously f(x) = 2005 for any given x if you choose f() suitably. (I used only one popular number theoretical function, but several times, and addition)

Alex

PS: and with eight 2s. Thanks, Euler!

[dave_dm]

You could express any number as

|{}|! + |{}|! + ... + |{}|!

where |{}|! = (cardinality of the empty set) factorial = 1

Without this kind of 'cheating' I can get 2005 down to seven 2's, I'd like to see Pauls two 2's solution though.

Dave

[ewmayer]

Originally Posted by xilman

I could manage 2005 with just four 4s and two 2s

Paul

Ah, but you're only allowed to use zeros as your building blocks (at least for the first problem), so unless you can express 2 with fewer zeros than two (e.g. 0!+0!) or 4 with fewer than 4, four 4s and two 2s translates to 4*4 + 2*2 = 20 zeros.

The first part of the problem statement:

"Use arbitrary mathematical symbols (+, -, *, /, ^, (, ), !, etc), functions & anything..."

seems to allow a lot of leeway, but any "function" would either have to be part of standard mathematical repertoire or defined using just zeros (whatever that means), so I stuck to basic arithmetic operators, i.e. just the ones in the above list.

Alex, if you really can do it with eight 2s, that would translate to 16 zeros, and earn you a "sehr gut." Being a fan of Fermat numbers, I was happy to settle for 17 zeros. (How's that for a whopper of a rationalization? )

[Wacky]

Well, I just sent in an answer with 14 zeros.

[xilman]

Originally Posted by Uncwilly

I cannot conceive how two 2's with any combination of operators can yield 2005.
Paul, the way that I understand it is to use only a selected digit (as many times as you wish), but only that digit.


Kemik: This thread: http://www.mersenneforum.org/showthread.php?t=2007 might be of help.
Also, notice that x/x = 1, which is often handy in these kinds of things.

I retract my claim.

I can do it by using a single 2 and a single 4.

For brevity, I will represent square root by S. Standard mathematical notation can be regained by a trivial combination. So, sqrt(n) is represented by S(n). lg is the standard binary logarithm function.

Evaluate the following series of expressions:
-lg(lg(S(2))
-lg(lg(SS2))
-lg(lg(SSS(2))
-lg(lg(SSSS(2))
-lg(lg(SSSSS(2))
-lg(lg(SSSSSS(2))
-lg(lg(SSSSSSS(2))
-lg(lg(SSSSSSSS(2))
-lg(lg(SSSSSSSSS(2))
-lg(lg(SSSSSSSSSS(2))
-lg(lg(SSSSSSSSSSS(2))
-lg(lg(SSSSSSSSSSSS(2))

How to represent 2005 should now be obvious.

As 2 = S(4), the solution which uses a single 4 should now also be obvious.


Paul

Addendum: if you don't like the use of lg, and I assure you it is a standard mathematical function, then I will use two more 2's or 4's and represent lg as log_2 or log_{S (4)} and make the base of the logarithms explicit. Not quite as elegant as the single digit solution, IMO, but should satisfy even the most pedantic.

[xilman]

Originally Posted by xilman

I retract my claim.

I can do it by using a single 2 and a single 4.

Note, further, that log(N)/log(sqrt(N)) = 2 for all N>0, and 2=0!+0!, so we never need more than two of any particular digit.

Paul

[Wacky]

Originally Posted by xilman

I can do it by using a single 2 and a single 4.

Perhaps you should have said "using a single 2 OR by using a single 4"

At worst, I think that you can certainly claim using no more than 3 instances of the digit 2 (to explicitly state the base of the log).

I hope that you have submitted these solutions.