document.write( "Question 1047141: Please help! I do not seem to understand the question. May be the question is wrongly put on exam. \r
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Algebra.Com's Answer #662674 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Something's wrong.  The first two sentences talk only about n,\r\n" );
document.write( "never mentioning k.  Then the last sentence asks about k, and\r\n" );
document.write( "doesn't even mention n.  I'll try to make something of it anyway.\r\n" );
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document.write( "I think the problem should have been:
When a positive integer n is divided by 5, the
\n" ); document.write( "remainder is 1. When n is divided 7, the remainder
\n" ); document.write( "is 3. What is the smallest possible positive integer n? 
If when n is divided by 5, the remainder is 1,\r\n" );
document.write( "then it is 1 more than a multiple of 5. So\r\n" );
document.write( "there is a positive integer p such that n = 5p+1\r\n" );
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document.write( "If when n is divided by 7, the remainder is 3,\r\n" );
document.write( "then it is 3 more than a multiple of 7.  So\r\n" );
document.write( "there is a positive integer q such that n = 7q+3\r\n" );
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document.write( "So n = 5p+1 = 7q+3\r\n" );
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document.write( "5p = 7q+2\r\n" );
document.write( "5p = 5q+2q+2\r\n" );
document.write( " p = q + (2/5)q + 2/5\r\n" );
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document.write( "p-q = (2/5)q + 2/5\r\n" );
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document.write( "The left side is an integer so the right side is too.\r\n" );
document.write( "Let that integer be A\r\n" );
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document.write( "p-q = (2/5)q + 2/5 = A\r\n" );
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document.write( "(2/5)q + 2/5 = A\r\n" );
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document.write( "2q + 2 = 5A\r\n" );
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document.write( "2q + 2 = 4A + A\r\n" );
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document.write( "q + 1 = 2A + A/2\r\n" );
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document.write( "q + 1 - 2A = A/2\r\n" );
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document.write( "The left side is an integer so the right side is too.\r\n" );
document.write( "Let that integer be B\r\n" );
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document.write( "q + 1 - 2A = A/2 = B\r\n" );
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document.write( "A/2 = B\r\n" );
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document.write( "A = 2B\r\n" );
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document.write( "q + 1 - 2A = B\r\n" );
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document.write( "q + 1 - 2(2B) = B\r\n" );
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document.write( "q + 1 - 4B = B\r\n" );
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document.write( "q + 1 = 5B\r\n" );
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document.write( "q = 5B-1\r\n" );
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document.write( "p-q = A\r\n" );
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document.write( "p - (5B-1) = 2B\r\n" );
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document.write( "p - 5B + 1 = 2B\r\n" );
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document.write( "p = 7B-1\r\n" );
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document.write( "n = 5p+1 = 7q+3\r\n" );
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document.write( "n = 5(7B-1)+1 = 7(5B-1)+3\r\n" );
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document.write( "n = 35B-5+1 = 35B-7+3\r\n" );
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document.write( "n = 35B-4 = 35B-4\r\n" );
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document.write( "4 is less than either 5 or 7, so let C = B-1, B = C+1\r\n" );
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document.write( "n = 35(C+1)-4\r\n" );
document.write( "n = 35C+35-4\r\n" );
document.write( "n = 35C+31\r\n" );
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document.write( "Since 35C divided by 5 or 7 there is 0 remainder. Therefore,\r\n" );
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document.write( "1. When we divide n by 5 we get the same remainder as when\r\n" );
document.write( "we divide 31 by 5, which gives a remainder of 1. \r\n" );
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document.write( "2. When we divide n by 7 we get the same remainder as when\r\n" );
document.write( "we divide 31 by 7, which gives a remainder of 3. \r\n" );
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document.write( "So the smallest integer that when divided by 5, the remainder \r\n" );
document.write( "is 1, and when divided by 7, the remainder is 3 is when C=0\r\n" );
document.write( "or n = 31. \r\n" );
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document.write( "That's what I think was the intended answer. \r\n" );
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document.write( "It would be impossible for n ever to be a multiple of 35, so\r\n" );
document.write( "the misprint (typo) could not have been that k was supposed\r\n" );
document.write( "to be n.\r\n" );
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document.write( "So the last sentence was probably from another problem, and\r\n" );
document.write( "somehow it got placed on the wrong problem.\r\n" );
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document.write( "Edwin
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