Given that a particular positive integer is a four-digit palindrome, what is the probability that it is a multiple of 99. Express your answer as a common fraction.
+199 I hope Rom will give you a detailed answer, but there are 10 possible choices out of 90 values, so 10/90=1/9.
+6251 there's probably some very clever way of attacking this but by the time you figure out what that is you'll have solved it by brute force and be off savoring a margarita.
First let's see what multiples of 99 are 4 digits. There can't be all that many of them.
99 x 11 = 1089 is the first one, 99 x 101 = 9999 is the last one, so 90 all together.
In this day and age of fast software I'd just code up a quick little program to see which of these are palindromes.
I come up with
{1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, 9999}
Now how many 4 digit palindromes are there. Well you're basically choosing 2 numbers.
The first digit can be 1-9, and the 2nd can be 0-9, so we've got 9x10 = 90 total palindromes. Thus
p = 10/90 = 1/9
