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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.

Guest Nov 4, 2018
 #1
avatar+147 
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I hope Rom will give you a detailed answer, but there are 10 possible choices out of 90 values, so 10/90=1/9. 

azsun  Nov 4, 2018
 #2
avatar+2786 
+3

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

Rom  Nov 4, 2018
edited by Rom  Nov 4, 2018

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