Find all powers of 2 such that, after deleting the first digit, another power of 2 remains. (For example, 2^{5} = 32. On deleting the initial 3, we are left with 2 = 2^{1}.) Numbers are written in standard decimal notation, with no leading zeroes.

Suppose the first digit of 2^{n} is *a*, and that after deleting that digit, we are left with 2^{m}.

Then 2^{n} = 10^{k}a + 2^{m}, for some integer k.

Hence 2^{n} − 2^{m} = 10^{k}a, and so 2^{n−m} − 1 = 2^{k−m}5^{k}a is divisible by 5.

If 2^{n−m} − 1 > 0 is divisible by 5, then n − m = 4r, for some positive integer r. (Consider the powers of 2, modulo 5.)

Hence 10^{k}a | = 2^{m}(2^{4r} − 1). |

= 2^{m}(2^{2r} + 1)(2^{2r} − 1). | |

= 2^{m}(2^{2r} + 1)(2^{r} + 1)(2^{r} − 1).(1) |

Note that, since 1 a 9, the left-hand side of (1) contains at most two distinct odd prime factors.

We will show that, if r > 1, the right-hand side of (1) must contain at least three distinct odd prime factors.

Note that 2^{2r} + 1 and 2^{2r} − 1 are odd integers that differ by 2. Hence they are relatively prime. (Their greatest common divisor must divide their difference, 2, and therefore must be equal to 1, as the integers are odd.)

Since 2^{2r} − 1 = (2^{r} + 1)(2^{r} − 1), 2^{2r} + 1 is relatively prime with 2^{r} + 1 and with 2^{r} − 1.

Similarly, 2^{r} + 1 and 2^{r} − 1 are odd integers that differ by 2, and are therefore relatively prime.

In conclusion, 2^{2r} + 1, 2^{r} + 1, and 2^{r} − 1 are all odd, pairwise relatively prime integers.

If r > 1, all of the integers are greater than 1, and hence each must have a distinct odd prime factor.

This contradicts the fact, noted above, that the left-hand side of (1) has at most two distinct odd prime factors.

We conclude that, for any solution, r = 1.

If r = 1, then 10^{k}a = 2^{m} · 3 · 5.

Hence k = 1, and a = 2^{m−1} · 3.

Since a 9, we must have (m, a) = (1, 3) or (2, 6).

Therefore, the only powers of 2 such that, on deleting the first digit, another power of 2 remains, are 2^{5} = 32 and 2^{6} = 64.

Source: Traditional