Surely the leading (= left-most) digit of a positive integer is an obvious thing? Just stare at the integer (e.g. 7823) and observe the left-most digit (7, and in this example)?

Suppose, clinic however, that you wanted to find the leading digit of a very large list of positive integers, a list so large it was hard to impossible to peruse by eye? How could you write an algorithm to compute the leading digits? Even more, suppose you wanted to come up with a mathematical argument that involved determining the leading digit of an otherwise unspecified positive integer?

In a short and lovely mathematical argument, Dave Radcliffe (@daveinstpaul) proves that there are exactly 18266 distinct ordered lists of values

(leading digit of 2ⁿ, … , leading digit of 9ⁿ)

as n ranges over the infinite set of positive integers.

A key part of his argument is that the leading digit of aⁿ is completely determined by the fractional part of n×log₁₀(a).

How might we see this?

Let’s make a table of values and see if something pops out:

k	fractional part of log10(k)
1	0.
2	0.30103
3	0.477121
4	0.60206
5	0.69897
6	0.778151
7	0.845098
8	0.90309
9	0.954243
10	0.
11	0.0413927
12	0.0791812
13	0.113943
14	0.146128
15	0.176091
16	0.20412
17	0.230449
18	0.255273
19	0.278754
20	0.30103

Nothing obvious, so what does Dave Radcliffe mean by “the leading digit of a_n is completely determined by the fractional part of n×log₁₀(a)”?

Let’s think about how we can algorithmically determine the leading digit of an integer written base 10.

Suppose k is a positive integer that is of the form a_pa_p-1…a₁a₀ base 10.

That is, the a_i are digits base 10 (i.e. one of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and a_p is not 0, because it is the leading digit.

What every school child does not immediately recall is that this means

k = a_p10^p + a_p-110^p-1 +… + a₁10 + a₀

So 10^p is no bigger than k, which in turn is less than 10^p+1 :

10^p ≤ k < 10^p+1

Therefore,

log₁₀(10^p ) ≤ log₁₀(k) < log₁₀(10^p+1)

because the logarithm is an increasing function.

In other words,

p ≤ log₁₀(k) < p+1

 which means that p is the greatest integer less than or equal to log₁₀(k)  – that is the floor of log₁₀(k): p=Floor[log₁₀(k)].

Now if we divide k by 10^p  we get:

k/10^p = a_p + 0.a_p-…a₁a₀ (base 10)

which means a_p = Floor[k/10^p] = Floor[k/10 ^Floor[log₁₀^(k)] ]

We can express Floor[log₁₀(k)] in terms of the fractional part { log₁₀(k)} of log₁₀(k) simply as

 Floor[log₁₀(k)] = log₁₀(k) – { log₁₀(k)}

So,

10 ^Floor[log₁₀^(k)] = 10^log₁₀^{(k) – {log}₁₀^(k}} = k/10^{log₁₀^(k}}

which, upon substituting into the expression above for a_p, gives:

a_p =Floor[10^{log₁₀^(k}}]

This is the precise sense in which the leading digit, a_p, of k is determined by the fractional part {log₁₀(k} of log₁₀(k).

When k= aⁿ, this is just the fractional part of n×log₁₀(a).

Going back to the table above, and including a third column of Floor[10^{log₁₀^(k}}], we get:

k	fractional part of log10(k)	Floor[10{log10(k}}]
1	0.	1
2	0.30103	2
3	0.477121	3
4	0.60206	4
5	0.69897	5
6	0.778151	6
7	0.845098	7
8	0.90309	8
9	0.954243	9
10	0.	1
11	0.0413927	1
12	0.0791812	1
13	0.113943	1
14	0.146128	1
15	0.176091	1
16	0.20412	1
17	0.230449	1
18	0.255273	1
19	0.278754	1
20	0.30103	2

Or, if we should do this for k = 2ⁿ for varying n, we get a table that begins as follows:

n	2n	fractional part of n ×log10(2)	Floor[10{n×log10(2}}]
1	2	0.30103	2
2	4	0.60206	4
3	8	0.90309	8
4	16	0.20412	1
5	32	0.50515	3
6	64	0.80618	6
7	128	0.10721	1
8	256	0.40824	2
9	512	0.70927	5
10	1024	0.0103	1
11	2048	0.31133	2
12	4096	0.61236	4
13	8192	0.91339	8
14	16384	0.21442	1
15	32768	0.51545	3
16	65536	0.81648	6
17	131072	0.11751	1
18	262144	0.41854	2
19	524288	0.71957	5
20	1048576	0.0205999	1

in precise agreement with Dave Radcliffe’s assertion.