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.

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1991 AHSME Problems/Problem 17

Problem

Solution

See also