document.write( "Question 1047141: Please help! I do not seem to understand the question. May be the question is wrongly put on exam. \r
\n" ); document.write( "\n" ); document.write( "When a positive integer n is divided by 5, the remainder is 1. When n is divided 7, the remainder is 3. What is the smallest positive integer k such that is a multiple of 35?
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Algebra.Com's Answer #662675 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "The number \"n\" is such that it has to satisfy these two series
\n" ); document.write( "a. \"5m+%2B+1\"
\n" ); document.write( "b. \"7j+%2B+3\"\r
\n" ); document.write( "\n" ); document.write( "To solve these types first take the \"LCM\" of the multiples \"5\" and {{7}}} here = \"35\".\r
\n" ); document.write( "\n" ); document.write( "So the resulting series will be:
\n" ); document.write( "\"35i+%2B+r\"\r
\n" ); document.write( "\n" ); document.write( "Now we find \"r\". To do this set \"i=0\" and try to find the first term in the series \"a\" and \"b\" that is \"same\".\r
\n" ); document.write( "\n" ); document.write( "series a. \"6\",\"11\",\"16\",\"21\",\"26\",\"highlight%2831%29\"
\n" ); document.write( "series b. \"10\",\"17\",\"24\",\"highlight%2831%29\",\"38\"\r
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\n" ); document.write( "\n" ); document.write( " So it matches at \"highlight%2831%29\". So \"r=31\"\r
\n" ); document.write( "\n" ); document.write( "Therefore the resulting series is :
\n" ); document.write( "\"35i+%2B+31\" \r
\n" ); document.write( "\n" ); document.write( "So the smallest number which needs to be added to this is \"highlight%284%29\" to make it divisible by \"35\".\r
\n" ); document.write( "\n" ); document.write( "Therefore, answer is \"k=4+\"\r
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