AMC 12 Mock Exam 5 Problem 25

The key fact is $10^3=1001-1\equiv-1\pmod{1001}$, so $10^4\equiv-10$, $10^5\equiv-100$, and $10^6\equiv1$. A palindrome divisible by $1001$ is at least $1001$, so it has $4$ to $7$ digits. Below, $a\in\{1,\dots,9\}$ and the other letters are digits.

4 digits, $\overline{abba}=a(10^3+1)+b(10^2+10)\equiv110b$. Since $\gcd(110,1001)=11$, we need $91\mid b$, so $b=0$: $9$ palindromes ($1001,2002,\dots,9009$).

5 digits, $\overline{abcba}=a(10^4+1)+b(10^3+10)+100c\equiv-9a+9b+100c$. This lies between $-81$ and $981$, so it must equal $0$: $100c=9(a-b)$. Then $9\mid c$; $c=9$ would need $a-b=100$, so $c=0$ and $b=a$: $9$ palindromes.

6 digits, $\overline{abccba}\equiv a(-100+1)+b(-10+10)+c(-1+100)=99(c-a)$. Since $\gcd(99,1001)=11$, we need $91\mid c-a$, so $c=a$, with $b$ free: $9\cdot10=90$ palindromes.

7 digits, $\overline{abcdcba}=a(10^6+1)+b(10^5+10)+c(10^4+10^2)+d\cdot10^3\equiv2a-90b+90c-d$. This lies between $-817$ and $828$, so it must equal $0$: $2a-d=90(b-c)$. Since $|2a-d|\le18<90$, we need $b=c$ and $d=2a$, so $a\in\{1,2,3,4\}$ and $b$ is free: $4\cdot10=40$ palindromes.

The total is $9+9+90+40=\boxed{148}$.