Find \(1_6 + 2_6 + 3_6 + \cdots + 45_6\). Express your answer in base \(6\).
Thanks in advance!
+367 This problem is simply 1+2+3+4+5+6+7...
45 base 6 is 5*6^0+4*6^1 which is 29.
So 1+2+3+4+5+6+...29.
Using the finite arithmetic sequence sum n(a1+an)/2
29(1+29)/2
29*30/2
29*15
435
That is correct in base 10: You can list all the numbers in base 6 as follows:
(1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, 30, 31, 32, 33, 34, 35, 40, 41, 42, 43, 44, 45)=29 numbers.
Then sum them in base 10 as shown above =435 in base 10.
Convert 435 from base 10 to base 6:
435 / 6 = 3 as the remainder.
72 / 6 = 0 as the remainder.
12 / 6 = 0 as the remainder.
2/6 = 2 as the remainder.
so 435 in base 10 =2003(from the bottom up) in base 6, which is the answer to your question.
