document.write( "Question 13382: if 2x+1 is a multiple of 5,and if 2x+1<100, how many possible values of x are prime numbers \n" ); document.write( "
Algebra.Com's Answer #6759 by khwang(438)\"\" \"About 
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[qoute]
\n" ); document.write( " if 2x+1 is a multiple of 5,and if 2x+1<100, how many possible values of x are prime numbers [/quote]\r
\n" ); document.write( "\n" ); document.write( " Sol: Note2x+1 is odd, if 2x+1 = 5q for some positive integer q.
\n" ); document.write( " then q must be odd. We can set q = 2k -1 for some psitive integer k.
\n" ); document.write( " Hence,if 2x+1 = 5(2k-1) = 10 k -5 ,then x = 5k -3. (why?)
\n" ); document.write( " Since 2x+1 <100 and x ( >0)is prime , so x = 5k -3 <= 49 or 5k <= 52.
\n" ); document.write( " k start from 1, 2,4...,up to 10 as
\n" ); document.write( " (to make x to be prime,k is not odd >1 ormultiple of3)
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\n" ); document.write( " k | x = 5k-3
\n" ); document.write( " ---------------
\n" ); document.write( " 1 | 2 (prime(OK)
\n" ); document.write( " 2 | 7 (prime(OK)
\n" ); document.write( " 4 | 17 (prime(OK)
\n" ); document.write( " 8 | 37 (prime(OK)
\n" ); document.write( " 10 | 47 (prime(OK)\r
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\n" ); document.write( "\n" ); document.write( " Hence, there are 5 such primes x with 0< 2x+1 < 100 and (2x+1) = 0 mod 5.\r
\n" ); document.write( "\n" ); document.write( " The above solution is the shortest way to get the answer withour redundant
\n" ); document.write( " testing.
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\n" ); document.write( " Sorry, I won't give further explanations.
\n" ); document.write( " Try to read carefully to understand the details.\r
\n" ); document.write( "\n" ); document.write( " Kenny
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