# 1991 AHSME Problems/Problem 17

## Problem

A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$. The year 1991 is the only year in the current century with the following 2 properties:

(a) It is a palindrome (b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrome.

How many years in the millenium between 1000 and 2000 have properties (a) and (b)? $\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

## Solution

Solution by e_power_pi_times_i

Notice that all four-digit palindromes are divisible by $11$, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$, which also means that the last digit of the three-digit number is $1$. Checking through the three-digit numbers $101, 111, 121,\dots, 181$, we find out that there are $\boxed{\textbf{(D) } 4}$ three-digit prime numbers, which when multiplied by $11$, result in palindromes.

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