As Easy As 1234
Problem
Using each of the digits 1, 2, 3, and 4, once and only once, with the basic rules of arithmetic (+, –, , ÷, and parentheses), express all of the integers from 1 to 25.
For example, 1 = 2 3 – (1 + 4)
Solution
Of course, there are may be other ways of arriving at each of these numbers:

Extensions
 If you are now permitted to use square roots, exponents, and factorials, can you produce all of the integers from 1 to 100?
 What is the first natural number that cannot be derived?
 Which is the first number that cannot be obtained if you are only permitted to use the basic rules of arithmetic (+, –, , and ÷)?
 What is the largest known prime you can produce?
Notes
Surprisingly it is possible to produce any finite integer using logarithms in a rather ingenious way.
We can see that,
2 = 2^{1/2}
2 = (2^{1/2})^{1/2} = 2^{1/4}
2 = ((2^{1/2})^{1/2})^{1/2} = 2^{1/8}, and so on.
2 = (2^{1/2})^{1/2} = 2^{1/4}
2 = ((2^{1/2})^{1/2})^{1/2} = 2^{1/8}, and so on.
Therefore,
log_{2}(2) = 1/2 = (1/2)^{1}
log_{2}(2) = 1/4 = (1/2)^{2}
log_{2}(2) = 1/8 = (1/2)^{3}, et cetera.
log_{2}(2) = 1/4 = (1/2)^{2}
log_{2}(2) = 1/8 = (1/2)^{3}, et cetera.
Hence,
log_{1/2}(log_{2}(2)) = 1
log_{1/2}(log_{2}(2)) = 2
log_{1/2}(log_{2}(2)) = 3, ...
log_{1/2}(log_{2}(2)) = 2
log_{1/2}(log_{2}(2)) = 3, ...
By using the integer part function, 1/[(3!)] = 1/[2.449...] = 1/2, we can obtain the required base 1/2, and using 4 to obtain the base 2, we can now produce any finite integer using the digits 1, 2, 3, and 4.
log_{(1/[(3!)])}(log_{4}(2)) = 1
log_{(1/[(3!)])}(log_{4}(2)) = 2
log_{(1/[(3!)])}(log_{4}(2)) = 3, ...
Problem ID: 13 (Sep 2000) Difficulty: 1 Star