262,083 views
14 votes
14 votes
Can someone answer this? I'll mark u the brialintest!

Can someone answer this? I'll mark u the brialintest!-example-1
User Asif Rahaman
by
3.1k points

1 Answer

21 votes
21 votes

pre convert single digits, or sometimes parts of a sum, then sum up. like this:

1.) 2³+2²+2^1+2^0 = 8+4+0+1 = 13

2.) 512 + 0 + 128 + 0 + 0 + 16 + 0 + 4 + 2 + 0 = 1010010110

you just need to know what the powers of two are, then you know how to convert numbers.

like 100 = 4 and 1 = 1 then 101 = 4 + 1 and in the other direction just take chunks of 5 until theres nothing left, but care not to overshoot the sum

5 = 2² + 0 + 2^1 = 101 in base2

the zeros are for the order, every time I would overshoot, it's a zero.

3.) 16 + 2 + 19 = 37

4.) 8² is 64, or (100)base8

8³ or 1000 would already overshoot, so let's see what we can do with the smaller value of a digit in base8, 8² = 64

we can 8 different digits (0 to 7) in base8

7x8^2 + 2*8^1 + 5*8^0

448 + 16 + 5

= (725)base8

very low battery right now. let me now if this already helps to grasp the concept. I'll come back later.

edit1: recharging my phone now. but I need to show quickly in order to get my own stuff done.

quick side note on topic: the number always stays the same, like if there would be a number of beans lying on the table. we justcrrorder them to write in another system. in normal base 10, 312 beans would be grouped intuitively in 3*100, 1*10 and 1*1

or: 3*10^2 + 1*10^1 + 2*10^0

note that something to the power of zero is always one (e.g.: 5^0 is like 5÷5)

User Juno
by
2.8k points