I had an attempt at finding a palindromic cube a couple of years ago but with no success. I recently dusted it off and tried again still with no success. I then thought I'd try to analyse what my chances were. The following summarises what I found. I think the basic maths is OK, but there may be some arithmetic errors. I first looked at odd palindromic squares. I am ignoring solutions of a special form eg those with just ones and zeroes and assuming the problem is effectively random. Starting with a specific example I looked at 17 digit numbers. The range of such numbers is from 10^16 to 10^17 - 1. I ignored the minus 1 for simplicity. Square roots ending in zero are not applicable in this case so only 0.9 of roots are valid. Thus the number of square roots in the range is approximately 0.9 * (sqrt(10^17) - sqrt(10^16)) = 1.95 * 10^8 The number of palindromic numbers in the target range is 10^9, but only those ending in 1, 4, 5, 6 and 9 are of interest so there are 0.5 * 10^9 candidates. So the probability that a given perfect square is also palindromic is 0.5 * 10^9/(10^17 - 10^16) which is approximately 5.56 * 10^-9. Thus the average density of palindromic squares is 5.56 * 10^-9 * 1.95 * 10^8 = 1.084 It seems to assume the distribution of palindromic squares is a Poissin distribution which gives the probability of zero = exp(-1.0725) = .342 so the probability of 1 or more in the range is approximately .658. It is quite easy to generalise this to for any odd digit range 2n + 1. The range of numbers is 10^2n to 10^(2n + 1) - 1 The number of perfect squares is 0.9 * (sqrt(10^(2n + 1)) - sqrt(10^2n)) = 0.9 * sqrt(10^2n) * (sqrt(10) - 1) = 1.95 * 10^n The number of palindromes is 0.5 * 10^(n + 1). Thus the density of palindromic squares is 0.5 * 10^(n + 1) * 1.95 * 10^n / (10^(2n + 1) - 10^2n) = 0.5 * 10 * 1.95 * 10^2n / 9 * 10^2n = 1.084 and the probability of 1 or more palindromic squares in the range is .658 . This produces the interesting result that the probability of finding a palindromic square of any specific length is constant. Of course they get more difficult to find as the number of values that have to be tested increases as the length of the number increases. A similar result can be obtained for even length palindromic squares. In this case I'll go straight to the general case. For numbers with 2n digits the range is 10^(2n - 1) to 10^(2n) - 1 The number of perfect squares is 0.9 * (sqrt(10^(2n)) - sqrt(10^2n - 1)) = 0.9 * sqrt(10^2n) * (1 - sqrt(0.1)) = .615 * 10^n The number of palindromes is 0.5 * 10^(n). Thus the density of palindromic squares is 0.5 * 10^(n) * .615 * 10^n / (10^(2n) - 10^(2n - 1)) = 0.5 * 0.615 / .9 = .710 and the probability of 1 or more palindromic squares in the range is .508 . So an even-digit palindromic square is less probable than an odd-digit palindromic square. This is not surprising because with odd digit palindromes the middle digit is effectively a "free" digit which can take any value from 0 to 9. Now on to palindromic cubes. Once again I'll go straight to the general case starting with odd-digit palindromes. In this case palindromes can end in any digit except 0. The range of numbers is 10^2n to 10^(2n + 1) - 1 The number of perfect cubes is 0.9 * (cuberoot(10^(2n + 1)) - cuberoot(10^2n)) = 0.9 * 10^(2n/3) * (cuberoot(10) - 1) = 1.039 * 10^(2n/3) The number of palindromes is 0.9 * 10^(n + 1). Thus the density of palindromic cubes is 1.039 * 10^(2n/3) * 0.9 * 10^(n + 1) / (10^(2n + 1) - 10^2n) = (1.039 * 0.9 / 9) * 10^(2n/3) * 10^(n + 1) * 10^(-2n) = .1039 * 10^(1 - n/3) Notice this depends on 'n', so the density of possible solutions decreases as numbers get larger. Thus for n = 8 ie length 17 we have density = .00223 and the probabilty of 1 or more palindromic cubes in the range is 0.022 . For n = 20 ie length 41 we have density = 2.24 * 10^-7 and the probabilty of 1 or more palindromic cubes in the range is 2.24 * 10^-7 . The conclusion is that the probability of finding a palindromic cube is remote and the situation gets worse for higher powers.

Mike Bennett (email) [May 21, 2000].