Difference between revisions of "1991 AHSME Problems/Problem 17"
m (→Solution) |
m (→Problem) |
||
Line 4: | Line 4: | ||
(a) It is a palindrome | (a) It is a palindrome | ||
− | |||
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime 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)? | How many years in the millenium between 1000 and 2000 have properties (a) and (b)? | ||
+ | |||
+ | <math>\text{(A) } 1\quad | ||
+ | \text{(B) } 2\quad | ||
+ | \text{(C) } 3\quad | ||
+ | \text{(D) } 4\quad | ||
+ | \text{(E) } 5</math> | ||
== Solution == | == Solution == |
Revision as of 15:32, 28 September 2014
Problem
A positive integer is a palindrome if the integer obtained by reversing the sequence of digits of is equal to . 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)?
Solution
See also
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.