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Question 1129772: How many positive integers k less than 289 are there such that k^2+6k+7 is divisible by 289?
Found 4 solutions by math_helper, kohkche, greenestamps, ikleyn:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! I found 2, using brute force.
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perl -e 'for($i=1;$i<289;$i++) { $v = $i*$i + 6*$i +7; if ($v % 289 == 0) {print "$i\n";} } '
42
241
Answer by kohkche(3)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
I don't know of any easy way to find the answer algebraically....
You can of course ignore the answer from the other tutor. It was the first answer on this site from him, so it was probably just some joker who thought it would be fun to respond to your question with a useless answer.
Obviously there are not 331 positive integers less than 289....
You can use an excel spreadsheet or the table function of a graphing calculator to divide the expression by 289 for every integer from 1 to 288; you will find it is divisible by 4 positive integers less than 289: 42, 241, 285, and 287.
Answer by ikleyn(52790)  (Show Source):
You can put this solution on YOUR website! .
I checked it using Excel spreadsheet.
The answer is: there are 2 ( two, TWO) such positive integer numbers.
They are 42 and 241.
There are no other solutions.
So, I confirm the answer by the tutor @Math_helper.
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