Question

Could someone please help me solve this question?

Answer 1

so in base 8 te 1st number is 1 and in base 10 it has the same value

the last number in base 8 is 100 the base 10 equivalent is

\[1 \times 8^2 + 0 \times 8^1 + 0 \times 8^0\]

so then 100 base 8 is equal to 64 base 10.

so you are summing the numbers 1 to 64

so the sum of an arithmetic sequence is

\[S_{n} = \frac{n}{2}[ a + l]\]

the 1st number a = 1

the last number l = 64

and there are 64 numbers

substitute them into the formula above for find the sum.

Hope this helps.

Answer 2

Get some paper and write the numbers in rows in the manner below:

0 1 2 3 4 5 6 7
10 11 12 13 14 15 16 17
20 ......
30 ......
.
.
.
.
.
110 111 112 113 114 115 116 117
140 141 142 143 144

The units digits in each of the rows (but not the last row) add up to 28.

The tens digits in each of the first 8 rows add up to 28.

So, the sum of the digits in the first eight rows is 8*28 + 8*28 which is 448 in base ten.

Use the same type thinking for the next four rows to get that sum in base ten. @Sepeario

To that sum, add the sum of the digits in the last row. That will be 5 + 6 + 7 + 8 + 9 = ? in base ten.

Then add all three sums to get the final answer. If you post it here, someone will check it.