Need help with a math question.?

I used this site to help

http://fclass.vaniercollege.qc.ca/web/mathematics/...

2010 in base 2 is 11111011010

I entered numbers into the calculator at the above site until I got the next 2 numbers

2012=11111011100

2019=11111100011

First figure out what 2010 is in binary notation. They tell you there are 11 digits

remember 2^11= 2048 2^10 = 1024 2^9 =512 2^8 =256 2^7 =128 2^6 =64 ...

2010 is 2048 which in binary is 100,000,000,000 (12 digits) and the question tells us there are 11 digits

so

10 9 8 7 6 5 4 3 2 1 0

1024 512 256 128 64 32 16 8 4 2 1

1 1 1 1 1 1 1 1 1 1 1 = 2047 or 2^12-1 = 2048-1

2047-2010 = 37 = 32+4+1 so 2010 is 11,111,011,010 i,e, remove the 32,4 and 1 digit

Add 1 to the binary number ....010

The next 7 numbers are

11,111,011,011

11,111,011,100 3 zeros = 2012

11,111,011,101

11,111,011,110

11,111,011,111

11,111,100,000

11,111,100,001

11,111,100,010

11,111,100,011 3 zeros = 2019

Checking a)

11,111,011,111 = 2047-32=2015 = 2010 +5 YES!!!

Checking b)

start with low numbers

1=001

2=010 We started here so the numbers below are the rest of the sequence

3=011

4=100

5=101

6=110

7=111

then a big jump