Question 238190
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First find the smallest 3-digit number that is evenly divisible by both 5 and 6.  Then add 1.


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.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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