Difference between revisions of "1991 AHSME Problems/Problem 17"

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(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)?
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<math>\text{(A) } 1\quad
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\text{(B) } 2\quad
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\text{(C) } 3\quad
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\text{(D) } 4\quad
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\text{(E) } 5</math>
  
 
== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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Solution by e_power_pi_times_i
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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 ==
 
== See also ==

Latest revision as of 11:38, 13 December 2016

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.

See also

1991 AHSME (ProblemsAnswer KeyResources)
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
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