The first approximation would just count the relative frequency of perfect squares to numbers in the 19-digit range. As the derivative of x^2 is 2x, and the square root of a 19-digit number beginning with 7 is going to be about 2,500,000,000 (the best I can do without a calculator), the squares are going by at a rate equal to about 5,000,000,000 times that of their square roots. So the first approximation of the probability is 1/5,000,000,000.
However, the last digit being 9 increases the chances that you have a perfect square: 1/5 of perfect squares end in 9, twice that which you'd expect if the last digit were uniformly distributed over the digits, so at this level of sophistication, we increase the probability to 1/2,500,000,000.
At the next level of sophistication, 1/25 of perfect squares end in 49, which is four time what you'd expect, so instead of multiplying our original estimate by 2, we multiply by 4, and the probability is now estimated at 1/1,250,000,000.
Of course in order to determine that 1/25 of perfect squares end in 49, you'd need a table of squares of numbers from 1 to 100. Why anyone would have one lying around, I don't know, as a computer can produce one at any time; its only advantage is being allowed in this puzzle. (Yes, I cheated by producing it on the fly via computer.)
One could continue on, with probabilities of ending in 549, 249, 749, 449, 549 and 849, and allow for the equal probability that the hundreds position could be any one of the given spare digits, but that is beyond what I'd want to do without a computer.
Edited on March 23, 2004, 3:01 pm
Posted by Charlie
on 2004-03-23 14:44:45