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2 for all, and all for 2

Posted by Karl Sharman on July 14, 2014 – 12:52 am

As an example, 64 is 2^6. Interestingly, take away the first number (6) and you are left with 4, which is 2^2 – So…find all the powers of 2 such that, after deleting the first digit, another power of 2 remains. Base 10, no leading zeroes etc.

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  1. 1. Wizard of Oz Said:

    32

  2. 2. slavy Said:

    Yep, that’s the complete list: 32 and 64

  3. 3. Karl Sharman Said:

    I just edited this post…. but I started with 12, and 132 is an option. Sooooo….. is “32 and 64″ your final answer?

  4. 4. slavy Said:

    Hi, Karl. I don’t get your last post! If, by “first digit” you mean the leading one then I’m 100% sure that the list is just “32 and 64″ and I have a solid proof. How are 12 and 132 related to the problem? They are neither powers of 2 nor related in the right way…

  5. 5. Karl Sharman Said:

    Well Slavy….It goes a little like this….

    Below is my original proof that 32 and 64 are the correct answers.

    After this (fairly conclusive) proof, I realised that 132, 164,232…. fit the bill of “deleting the first digit, another power of 2 remains”. This meant that all the below was rubbish!

    Then I saw your post and thought – you poor misguided fool. I read the question again, and one more time for luck, and could see no fault with my revised thinking – eg 132 fits the bill.

    A few hours later my stupidity dawned on me… you will be pleased to know that I have given myself a clip round the back of the head, and will be serving a few years in Purgatory for not reading the question properly – Rookie Mistake!!!

    Proof:

    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^ka + 2^m, for some integer k.
    Hence 2^n – 2^m = 10^ka, and so 2^n-m − 1 = 2^k−-m5^ka 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^ka = 2^m(2^4r − 1).
    = 2^m(2^2r + 1)(2^2r − 1).
    = 2^m(2^2r + 1)(2^r + 1)(2^r − 1)x(1)
    Note that, since 1 is less than or equal to a which is 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 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^ka = 2^m x 3 x 5.
    Hence k = 1, and a = 2^m-1 x 3.
    Since a is less tha 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 2^5 = 32 and 2^ = 64.

  6. 6. slavy Said:

    Hi, Karl. This is exactly the proof I had in mind :)

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