Difference between revisions of "1991 AHSME Problems/Problem 17"
(Created page with "A positive integer <math>N</math> is a ''palindrome'' if the integer obtained by reversing the sequence of digits of <math>N</math> is equal to <math>N</math>. The year 1991 is t...") |
m (→Solution) |
||
| (7 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| + | == Problem == | ||
| + | |||
A positive integer <math>N</math> is a ''palindrome'' if the integer obtained by reversing the sequence of digits of <math>N</math> is equal to <math>N</math>. The year 1991 is the only year in the current century with the following 2 properties: | A positive integer <math>N</math> is a ''palindrome'' if the integer obtained by reversing the sequence of digits of <math>N</math> is equal to <math>N</math>. The year 1991 is the only year in the current century with the following 2 properties: | ||
(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 by e_power_pi_times_i | ||
| + | |||
| + | Notice that all four-digit palindromes are divisible by <math>11</math>, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between <math>1100</math> and <math>2000</math>, which also means that the last digit of the three-digit number is <math>1</math>. Checking through the three-digit numbers <math>101, 111, 121,\dots, 181</math>, we find out that there are <math>\boxed{\textbf{(D) } 4}</math> three-digit prime numbers, which when multiplied by <math>11</math>, result in palindromes. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1991|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:38, 13 December 2016
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
Solution by e_power_pi_times_i
Notice that all four-digit palindromes are divisible by 
, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between 
and 
, which also means that the last digit of the three-digit number is 
. Checking through the three-digit numbers 
, we find out that there are 
three-digit prime numbers, which when multiplied by , result in palindromes.
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. 
