Saturday, March 23, 2002
M I N D  G A M E S

Buddha is smiling

THANK you for your tremendous response to Mind Games. Letters have been coming in regularly, but I had not anticipated the mailbag to be so big. I shall get back to each one of you individually, but you'll have to give me one more week for that. It is a pleasure to know that some of you have developed interest in mathematics after reading this column, which is its intended goal.

To Thales the primary question was not what do we know, but how do we know it.


Most of you want to know what book I refer to for coming up with a problem every week; so, the answer is that there is no such book. I remember having written in this column that the book of all answers is a myth from schooldays and contrary to the true nature of mathematics. One should read the works of masters and not their pupils (By masters, I mean greats like Newton, Ramanujan and Fibonacci).

Solving the problem from the previous week - of three numbers raised to a certain power each - requires that you have no calculator or computer. Ravinder Mittal, Rajeev Kumar Tak and Amandeep Jindal came up with the correct answer, but they did not tell me how did they arrive at it. Here's how the solution has been explained in an ancient text, from the time when there was no computer: First, for the three numbers A, B, and C, let's define the function floor(x), where x is a real number, such that floor(x) = the integer part of x. Let y = floor(loga (x)) + 1 . As a general rule, y will be the number of digits of x in base a. If we reverse this, we can say that x is somewhere between a y and ay + 1. Another basic rule is loga (bc) = cloga (b). If we don't use this rule, the calculation cannot be handled using any standard scientific calculators, as these can't handle calculation with numbers greater than 10,100. If we use these two rules to A, B and C in base 10, it will show that A has 6,609 digits, B has 6,606 digits, and C has 6,603 digits in base 10. Therefore, A is bigger than B which in turn is bigger than C. A is the biggest, and C is the smallest.


A similar solution uses the fact that the logarithm function is an increasing function, so, it follows that logA > logB if and only if A > B. Hence, logA = 2002 log2000 " 2002(3.010) " 6608.662; logB = 2001 log2001 " 6605.795; logC = 2000 log2002 " 6602.928. The approximate difference is given by : logA - logB = logA/B " 3, hence, A " 103B. Similarly, B " 103C. Thus A > B > C. A more general solution is: Claim: A > B > C; proof: A > B if and only if logA > logB. I shall prove logA - logB > 0 i.e. 2002log2000 - 2001log2001 > 0. Let f(x) = (x + 2)logx - (x+1)log(x+1) so that, for example, f(2) = 4log2 - 3log3. Differentiating this function, f(x) = (x + 2)/x + logx - 1 - log(x + 1) = 2/x -log[(x+1)/x]. This derivative is positive if and only if e2/x > (x+1)/x. Using ey > 1 + y for all y, let y = 2/x. We have e2/x > 1 + 2/x = (x + 2)/x > (x + 1)/x. So the function f is increasing, in particular, f(2000) = logA - logB > 0 and it follows that A > B. The proof that B > C is similar.

There are many ways to solving a problem, which each solver may choose according to his or her personality. We'll talk more, till then, keep writing in at The Tribune or

Aditya Rishi