A few days ago I mentioned a passing comment in video by Richard Boucherds. This post picks up on another off-hand remark from that post.

Boucherds was discussing why exp(π √67) and exp(π √163) are nearly an integer.

exp(π √67) = 147197952744 – ε_{1}

exp(π √163) = 262537412640768744 – ε_{2}

where ε_{1} and ε_{2} and on the order of 10^{-6} and 10^{-12} respectively.

He called attention to the 744 at the end and commented that this isn’t just an artifact of base 10. In many other bases, these numbers end in that bases’ representation of 744. This is what I wanted to expand on.

To illustrate Boucherds’ remark in hexadecimal, note that

147197952744 -> 0x2245ae82e8 262537412640768744 -> 0x3a4b862c4b402e8 744 -> 0x2e8

Boucherds’ comment is equivalent to saying

147197952744 = 744 mod *m*

and

262537412640768744 = 744 mod *m*

for many values of *m*. Equivalently 147197952000 and 262537412640768000 have a lot of factors; every factor of these numbers is a base where the congruence holds.

So for how many bases *m* are these numbers congruent to 744?

The number of factors of a number *n* is written *d*(*n*). This is a multiplicative function, meaning that for relatively prime numbers *a* and *b*,

*d*(*ab*) = *d*(*a*) *d*(*b*).

Note that

147197952000 = 2^{15} 3^{3} 5^{3} 11^{3}

and

262537412640768000 = 2^{18} 3^{3} 5^{3} 23^{3} 29^{3}

It follows that

*d*(147197952000) =

*d*(2^{15} 3^{3} 5^{3} 11^{3}) =

*d*(2^{15}) *d*(3^{3}) *d*(5^{3}) *d*(11^{3}).

Now for any prime power *p*^{k}

*d*(*p*^{k}) = *k* + 1,

and so

*d*(147197952000) = 16 × 4 × 4 × 4.

Similarly

*d*(262537412640768000) = 19 × 4 × 4 × 4 × 4.

For more about almost integers, watch Boucherds’ video and see this post.