You have created a 19 digit number with your 20 digit tiles as follows:

7_340_46_2010_51_49

Unfortunately someone knocked out 5 of the number tiles and placed them with the remaining number tile. The 6 tiles that are out are 6 3 2 8 9 3.

Without using any calculators, programs, or similar devices, what is the easiest way to figure out what the probability that my original 19-digit number will be a perfect square and what is this probability?

Note: The pesky "someone" also renumbered the number tiles so they were wrong and you couldn't tell what the right numbers are. This has been fixed to make the problem easier.

My first thoughts had to do with the one's place digit.

The one's place of any square number is as follows, in the form of One's digit of X : One's digit of X^2

0:0   1:1   2:4   3:9   4:6   5:5   6:6   7:9   8:4   9:1

So, out of these 10 possibilities, a 9 for the squared number shows up twice.  So how about the answer being a 1/5 probability of the 19 digit number being a perfect square?

This is probably too simple and naive of an approach, but I thought it was possible that all the other information was supposed to be distracting/misleading.  I figured this idea was worth embarassing myself =)

Otherwise, I like Charlie's approach with the 1(even digit)49 concept.

Posted by nikki on 2004-03-23 16:07:58