If you were asked to name a power of 2 that begins with a 3, no doubt you would quickly reply “32” (= 25). Again, the request for a power of 2 that begins with 12 would soon bring “128” (= 27). But if you were asked for a power of 2 that begins with 11223344556677, you would likely remain silent. Indeed, you may well wonder whether such a power of 2 exists. The remarkable theorem that we prove in this essay settles this question; it asserts that there exist powers of 2 beginning with any gizen sequence of digits. As a matter of fact, the theorem makes the same claim for 3, 4, and any positive integer a which is not a power of 10 (i.e., a ≠ 1, 10, 100, 1000, etc.). We prove the theorem for powers of 2; the general case is established in the same way.

Let S = abc … K be any sequence of digits. We need to show that, for some n,

2n = abc … k …

where there may be digits beyond the last digit of S.

What we want to consider first is the set of numbers which begin with the digits S. To fix the ideas, let us begin with a concrete example, say S = 5. (Here the required power is 29 = 512.) If 2n is to begin with a 5, then it must occur in one of the intervals

Every integer in these intervals begins with a 5, and no others do.