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Q: Where can you find detailed information about SGT John R. Nofsinger 333rd INF 84th DIV died 02Dec44?
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The Black Hawks' first owner was coffee tycoon Frederic McLaughlin. He had been a commander with the 333rd Machine Gun Battalion of the 86th Infantry Division during World War I. This Division was nicknamed the "Blackhawk Division", after a Native American of the Sauk nation, Chief Black Hawk, who was a prominent figure in the history of Illinois. McLaughlin evidently named the hockey team in honor of the military unit, making it one of many sports team names using Native Americans as icons. For many years, the name was spelled primarily "Black Hawks", but sometimes "Blackhawks", even by the club itself. This ambiguity was finally settled in the summer of 1986 when the club officially decided on the one-word version based on the spelling found in the original franchise documents.


What is the sum of all the multiples of 3 or 5 below 1000?

The first multiple of 3 is 3; the last multiple of 3 below 1000 is 999. (This can be determined by dividing 1000 by 3, ignoring the remainder, and then multiplying that number by 3 to determine the largest multiple of 3 less than 1000.) Since 999 is 3 times 333, there are 333 multiples of 3 that are less than 1000. So, consider them by pairs: 1st and 333rd = 3 + 999 = 1002 2nd and 332nd = (2 x 3) + (332 x 3) = 6 + 996 = 1002 3rd and 331st = (3 x 3) + (331 x 3) = 9 + 993 = 1002 up to 166th and 168th = (166 x 3) + (168 x 3) = 498 + 504 = 1002 167th = 501 (which is half of 1002) In other words, since 333 is not an even number, there are (333 - 1)/2 = 166 pairs, plus that extra half of a pair. The sum of all the multiples of 3 less than 1000 is 166.5 x 1002 = 166,833. The same can be done for the multiples of 5. The first multiple of 5 is 5; the last multiple of 5 below 1000 is 995. (This can be determined by dividing 1000 by 5, subtracting 1 since it divided evenly and you need the largest multiple less than 1000, and then multiplying that number by 5 to determine the largest multiple of 5 less than 1000.) Since 995 is 5 times 199, there are 199 multiples of 5 that are less than 1000. So, consider these by pairs as well.e are 1st and 199 = 5 + 995 = 1000 2nd and 198 = 10 + 990 = 1000 up to 100th = 500 (which is half of 1000) In other words, since 199 is not an even number, there are (199 - 1)/2 = 99 pairs, plus that extra half of a pair. So, the sum of all the multiples of 5 less than 1000 is 99.5 x 1000 = 99,500. If the desired answer is the sum of all numbers less than 1000 that are either multiples of 3 or 5, then the numbers that are multiples of both 3 and 5 have been included twice - once as multiples of 3 and again as multiples of 5. So, since all numbers that are both multiples of 3 and multiples of 5 are multiples of 15, determine the sum of all the multiples of 15 and subtract it from the sum of the multiples of 3 and the multiples of 5. We can repeat the same procedure again. The first multiple of 15 is 15; the last multiple of 15 less than 1000 is 990. Since 990 is 15 x 66, there are 66 multiples of 15 less than 1000. 1st and 66th = 15 + 990 = 1005 2nd and 65th = 30 + 975 = 1005 and so on Since 66 is an even number, there are 66/2 = 33 pairs. So the sum of all the multiples of 15 less than 1000 is 33 x 1005 = 33,165. The sum of all the multiples of 3 and all the multiples of 5, but not counting them twice, is 166,833 + 99,500 - 33,165 = 233,168.


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