bjuveges
New member
- Joined
- May 26, 2014
- Messages
- 13
[h=2]Problem 11[/h]A palindrome between
and
is chosen at random. What is the probability that it is divisible by
?

[h=2]Solution[/h]View the palindrome as some number with form (decimal representation):
. But because the number is a palindrome,
. Recombining this yields
. 1001 is divisible by 7, which means that as long as
, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (
) possibilities for palindromes. However, if
, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to 
Basically I only need help in understanding how the answer was found so:
1. How did they get the variables to begin this equation such as a^3,a^2,a^1,a^0 and 10^3,10^2,10?
2. 1001a^3+100a^2? the palindrome part I understand, but how they recombined the values confuses me.
3. a^2=0 & a^2=7 creating different possibilities confuses me.
If anything, I really wanted a better explanation of the problem, using the same method as the solution
and
is chosen at random. What is the probability that it is divisible by
?
[h=2]Solution[/h]View the palindrome as some number with form (decimal representation):
. But because the number is a palindrome,
. Recombining this yields
. 1001 is divisible by 7, which means that as long as
, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (
) possibilities for palindromes. However, if
, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to 
Basically I only need help in understanding how the answer was found so:
1. How did they get the variables to begin this equation such as a^3,a^2,a^1,a^0 and 10^3,10^2,10?
2. 1001a^3+100a^2? the palindrome part I understand, but how they recombined the values confuses me.
3. a^2=0 & a^2=7 creating different possibilities confuses me.
If anything, I really wanted a better explanation of the problem, using the same method as the solution