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Solution to puzzle 118: Powers of 2: deleted digit

Skip restatement of puzzle.Find all powers of 2 such that, after deleting the first digit, another power of 2 remains.  (For example, 25 = 32.  On deleting the initial 3, we are left with 2 = 21.)  Numbers are written in standard decimal notation, with no leading zeroes.


Suppose the first digit of 2n is a, and that after deleting that digit, we are left with 2m.
Then 2n = 10ka + 2m, for some integer k.
Hence 2n − 2m = 10ka, and so 2n−m − 1 = 2k−m5ka is divisible by 5.
If 2n−m − 1 > 0 is divisible by 5, then n − m = 4r, for some positive integer r.  (Consider the powers of 2, modulo 5.)

Hence 10ka = 2m(24r − 1).
 = 2m(22r + 1)(22r − 1).
 = 2m(22r + 1)(2r + 1)(2r − 1).(1)

Note that, since 1 less than or equal to a less than or equal to 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 22r + 1 and 22r − 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 22r − 1 = (2r + 1)(2r − 1), 22r + 1 is relatively prime with 2r + 1 and with 2r − 1.
Similarly, 2r + 1 and 2r − 1 are odd integers that differ by 2, and are therefore relatively prime.
In conclusion, 22r + 1, 2r + 1, and 2r − 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 10ka = 2m · 3 · 5.
Hence k = 1, and a = 2m−1 · 3.
Since a less than or equal to 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 25 = 32 and 26 = 64.

Source: Traditional

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