I talk here about how a mathematician mindset can be useful for AI alignment. But first, a puzzle:

Given m, what is the least number n≥2 such that for 2≤k≤m, the base k representation of n consists entirely of 0s and 1s?

If you want to think about it yourself, stop reading.

For m=2, n=2.

For m=3, n=3.

For m=4, n=4.

For m=5, n=82,000.

Indeed, 82,000 is 10100000001010000 in binary, 11011111001 in ternary, 110001100 in base 4, and 10111000 in base 5.

What about when m=6?

So, a mathematician might tell you that this is an open problem. It is not known if there is any n≥2 which consists of 0s and 1s in bases 2 through 6.

A scientist, on the other hand, might just tell you that clearly no such number exists. There are 2k−1 numbers that consist of k 0s and 1s in base 6. Each of these has roughly log5(6)⋅k digits in base 5, and assuming things are roughly evenly distributed, each of these digits is a 0 or a 1 with "probability" 2/5. The "probability" that there is any number of length k that has the property is thus less than 2k⋅(2/5)k=(4/5)k. This means that as you increase

I talk here about how a mathematician mindset can be useful for AI alignment. But first, a puzzle:

Given m, what is the least number n≥2 such that for 2≤k≤m, the base k representation of n consists entirely of 0s and 1s?

If you want to think about it yourself, stop reading.

For m=2, n=2.

For m=3, n=3.

For m=4, n=4.

For m=5, n=82,000.

Indeed, 82,000 is 10100000001010000 in binary, 11011111001 in ternary, 110001100 in base 4, and 10111000 in base 5.

What about when m=6?

So, a mathematician might tell you that this is an open problem. It is not known if there is any n≥2 which consists of 0s and 1s in bases 2 through 6.

A scientist, on the other hand, might just tell you that clearly no such number exists. There are 2k−1 numbers that consist of k 0s and 1s in base 6. Each of these has roughly log5(6)⋅k digits in base 5, and assuming things are roughly evenly distributed, each of these digits is a 0 or a 1 with "probability" 2/5. The "probability" that there is any number of length k that has the property is thus less than 2k⋅(2/5)k=(4/5)k. This means that as you increase