A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 8-digit palindrome $n$ is chosen uniformly at random. What is the probability that $\frac{n}{11}$ is also a palindrome?
$\textbf{(A)} \ \frac{209}{900} \qquad \textbf{(B)} \ \frac{143}{600} \qquad \textbf{(C)} \ \frac{6}{25} \qquad \textbf{(D)} \ \frac{77}{300} \qquad \textbf{(E)} \ \frac{119}{900}$
answer =B, 143/600
Start with a 7 -digit number $ABCDCBA$, multiply by 11, we get $A(A+B)(B+C)(C+D)(D+C)(C+B)(B+A)A$, to make this to be a palindrome, we need to have
$C+D<10, B+C<10, A+B<10$, just doing case work with $B$,
B= 0 , 495 numbers;
B= 1 , 432 numbers ;
B= 2 , 364 numbers;
B= 3 , 294 numbers;
B= 4 , 225 numbers;
B= 5 , 160 numbers;
B= 6 , 102 numbers;
B= 7 , 54 numbers;
B= 8 , 19 numbers;
B= 9 , 0
add up get $2145$, so the answer is 2145/9000=143/600
AMC 10/12 exam time:
A: Nov. 10, 2022
B: Nov 16, 2022
AMC 8: January, 17-23, 2023
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