Binary Cascade

Number Theory Level 1

If $A = 1001001_{(2)}$ , then $A + 2A + 4A + 1 = ?$

111111111_{(2)}

1000000000_{(2)}

1001001001_{(2)}

1111111111_{(2)}

3 solutions

David Sherratt
Feb 15, 2016

$A$ is odd (because last digit is $1$ ). $2A$ is even (since anything times 2 is even). Therefore $4A$ is even, and $1$ is odd.

odd + even + even + odd = even.

Only even option is $1000000000_{2}$ .

I was very lazy with my approach.

Or, you could say "efficient." ;-) I was hoping that someone would post this solution.

Zandra Vinegar Staff - 5 years, 4 months ago

Zandra Vinegar Staff
Feb 14, 2016

$A = 1001001_{(2)}$
$2A = 10010010_{(2)}$
$4A = 100100100_{(2)}$

Therefore, $A + 2A + 4A = 111111111_{(2)}$ and $A + 2A + 4A + 1 = 100000000_{(2)}$

Guillermo Templado
Feb 16, 2016

$A=1 + 8 + 64 = 73 ; 2A = 146 , 4A = 292 \Rightarrow$ $A + 2A + 4A + 1 = 512 = 2^9 = 1000000000_{(2)}$