I can count to 1100


#1

one = 0b1
two = 0b10
three = 0b11
four = 0b100
five = 0b101
six = 0b110
seven = 0b111
eight = 0b1000
nine = 0b1001
ten = 0b1010
eleven = 0b1011
twelve = 0b10000

i solved them all but the twelve had a problem anyway i was wondring if you can explain that so i can deal great numbers like 1m whatever


#2

the leftmost number is 2 ** 0 which is 1
the second left is 2 ** 1 which is 2
the third left is 2 ** 2 which is 4
if there is a number "0b111", it means 4 + 2 + 1 = 7
while "0b101" means 4 + 0 + 1 = 5

moreover, 1 byte is 8 bits, that is to be 0b00000000 ~ 0b11111111
there could be 128 different result in this situation
0b00000000 = 0
0b11111111 = 127 = 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 +2^0 = 64 + 32 +16 +8 + 4 + 2 + 1 = 127

just a collection of numbers, take it easy :smiley:

and btw, if you want to know how to get the binary number from decimal numbers, i'm willing to answer it!


#3

no i already understand now thanks for help and i hope you a great day


#4

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