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Series Q (1 Viewer)

Estel

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Joined
Nov 12, 2003
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HSC
2005
Sum of all integers between 1 and 100 is
(n(a+l))/2
=99(100)/2
=4950

Sum of all integers that are multiples of 6
ie 6,12,18... 96
=16(102)/2
=816

Therefore answer is 4950-816
=4134
 
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