Question 168611
Note: sum 1 to n = n(n+1)/2


First term that is a multiple of 6: 3132


Last term that is a multiple of 6: 51804



3132+3138+3144+...+51798+51804



6(522+527+533+...+8628+8634)



6(sum 1 to 8634 - sum 1 to 521) 



6(8634(8635)/2 - 521(522)/2)



6(37277295 - 135981)



6(37141314)



222847884




So the sum of all multiples of 6 that are strictly between 3128 and 51804 is 222,847,884