document.write( "Question 238190: If a three digit number is divided by 5 or 6,the remainder is 1 in each case. What is the least such three digit number? \n" ); document.write( "

Algebra.Com's Answer #175040 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "First find the smallest 3-digit number that is evenly divisible by both 5 and 6. Then add 1.\r
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\n" ); document.write( "\n" ); document.write( "Since 5 is prime and not a factor of 6, the smallest positive integer evenly divisible by both 5 and 6 must be 5 times 6 or 30. So the question becomes what is the smallest 3-digit number evenly divisible by 30? 1, 2, and 3 times 30 each produce a 2-digit number. Hence the smallest integer multiplier of 30 that produces a 3-digit product is 4, and the smallest 3-digit integer divisible by 30 is 120. Hence, 120 is the smallest 3-digit integer evenly divisible by both 5 and 6. And finally, 121 is the smallest 3-digit integer when divided by either 5 or 6 leaves a remainder of 1 in each case.\r
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