Difference between revisions of "1991 AHSME Problems/Problem 17"
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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: | ||
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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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| + | == Solution == | ||
| + | <math>\fbox{}</math> | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1991|num-b=16|num-a=18}} | ||
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| + | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:05, 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 | ||
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