Question

The binary equivalent to the product of the
first prime number is​

Answer 1

Answer:

Don’t confuse the value of a number with its representation. In that sense you might hope that prime numbers should work the same way and lucky for us, they do.

However.

In order to determine whether a number is divisible by a prime (a divisor) that divides 10k−1 , one could group the digits in blocks of k digits. An example, for k=1 , is the famous rule that a number is divisible by 3 if and only if the sum of the digits is divisible by 3 . In fact a more precise statement can be made, the remainder after division by 3 is equal in both cases (of the number and of the sum of the digits).

If you use base 2 similar rules can be constructed, using the expression 2k−1 .

For instance:

A number is divisible by 3 if the sum of groups of 2 consecutive bits (counting from the right hand side), is divisible by 3, just plug in k=2 into 2k−1 . There is a catch though, you must compute the sum of the groups of digits in binary as well.

This also means we have a very simple rule to determine when a number is divisible by 7 .

We have 112 is divisible by 310 .

But also 1012 , split into groups of 2 digits 12,012 and 12+012=112 .

We have 1112 is divisible by 710 .

But also 10011012 , split into groups of 3 digits 12,0012,1012 and 12+0012+1012=1112

You could also try to find the analogy of the following rule:

A number is divisible by 11 if the alternating sum of the digits is divisible by 11 .

To be specific: what division rule does exist for 3 , if one uses the base 2 representation?