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

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</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, 191</math>, we find out that <math>\boxed{\textbf{(D) 4}}</math> three-digit numbers, when multiplied by <math>11</math>, result in palindromes.
  
 
== See also ==
 
== See also ==

Revision as of 11:35, 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 $101, 111, 121,\dots, 191$, we find out that $\boxed{\textbf{(D) 4}}$ three-digit numbers, when multiplied by $11$, result in palindromes.

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AHSME Problems and Solutions

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