Problem:
A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6 -digit palindrome n is chosen uniformly at random. What is the probability that 11n is also a palindrome?
Answer Choices:
A. 258
B. 10033
C. 207
D. 259
E. 3011
Solution:
Let n be a 6 -digit palindrome, m=11n, and suppose m is a palindrome as well. First, if m is a 4 -digit number, then n=11m<11⋅105= 106+105. Thus the first and last digit of n is 1 . Thus the last digit of m is 1 and then the first digit of m must be 1 as well. Then m≤1991<2000 and n=11m<11⋅2000=22000, which is a contradiction. Therefore m is a 5 -digit number abcba. If a+b≤9 and b+c≤9, then there are no carries in the sum n=11m=abcba0+abcba; thus the digits of n in order are a,a+b,b+c,b+c, a+b, and a. Conversely, if a+b≥10, then the first digit of n is a+1 and the last digit a; and if a+b≤9 but b+c≥10, then the second digit of n is a+b+1 if a+b<9, or 0 if a+b=9, and the previous to last digit is a+b. In any case n is not a palindrome. Therefore n=11 m is a palindrome if and only if a+b≤9 and b+c≤9.
Thus the number of pairs (m,n) is equal to
b=0∑9c=0∑9−ba=1∑9−b1=b=0∑9(10−b)(9−b)
Letting j=10−b gives
j=1∑10j(j−1)=610⋅11⋅21−210⋅11=330
The total number of 6 -digit palindromes abccba is determined by 10 choices for each of b and c, and 9 choices for a, for a total of 9⋅102=900. Thus the required probability is 900330=3011.
The problems on this page are the property of the MAA's American Mathematics Competitions