Question 588799
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The only numbers that are divisible by 5 end in either 0 or 5.  Numbers where the sum of the digits is divisible by 3 are divisible by 3.


Since the desired number must be strictly greater than 200, then next number that ends in either 0 or 5 is 205.  But the sum of the digits of 205 is 7 which is not divisible by 3.  The next number that is divisible by 5 is 210.  The sum of the digits of 210 is 3 which is divisible by 3, hence 210 is divisible by 5 because it ends in 0 and 210 is divisible by 3 because its sum of digits is divisible by 3.  Not only did we find <i><b>a</b></i> number >200 divisible by both 3 and 5, but we found the smallest number >200 that is divisible by both 3 and 5.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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