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Math Help - Determine the smallest value

  1. #1
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    Determine the smallest value

    Troubles again
    Determine the smallest value of |20^m-9^n| ; m and n are positive integers.
    I created a graph of it and as far as I've noticed 11 is the smallest value.
    The thing is- how to prove it?
    I tried to play with divisibility by 11, but I didn't succeed.
    Any ideas?
    Regards,
    Lukasz
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  2. #2
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    Re: Determine the smallest value

    Quote Originally Posted by Lukaszm View Post
    Troubles again
    Determine the smallest value of |20^m-9^n| ; m and n are positive integers.
    I created a graph of it and as far as I've noticed 11 is the smallest value.
    The thing is- how to prove it?
    I tried to play with divisibility by 11, but I didn't succeed.
    Any ideas?
    Regards,
    Lukasz
    I belive that your answer is correct.

    we can write twenty as

    20^m=(9+11)^m

    If we expand using the binomial theorem we get

    (9+11)^m=9^m+\binom{m}{1}9^{m-1}\cdot 11+ \cdots +\binom{m}{m-1}9\cdot11^{m-1}+11^{m}=\sum_{i=0}^{m}\binom{m}{i}9^{m-i}11^i

    Now if we subtract we get

    \bigg| 9^m-9^n+\sum_{i=1}^{m}\binom{m}{i}9^{m-i}11^i\bigg|

    If we set m=n the first two terms reduce out.

    Now since all terms of the sum have the same sign we must minimize the sum. the value m=1 gives this minimum

    This should give you the conclustion that you are looking for.
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  3. #3
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    Re: Determine the smallest value

    Here's a proof that k = 11.

    Let k = 20^m - 9^n s.t. |k| is a min among such numbers, n, m \in \mathbb{Z}^+.

    Look at it modulo 10: k \equiv (0) - (-1)^n \equiv (-1)^{n+1} \mod 10.

    Thus |k| \in \{ 1, 9, 11, 19, 21, 29, 31, ...\}.

    Since 11 = 20^1 - 9^1, and |k| is defined to be minimal, conclude that |k| \in \{ 1, 9, 11 \}.

    But 9 can be excluded, for if |k| = 9, then 20^m - 9^n = \pm 9, which is impossible (look \mod 9, recall n >0).

    Thus |k| \in \{ 1, 11 \}.

    Also, you can exclude k = 1, because:

    Assume k=1. Then 20^m - 9^n = 1, which implies 20^m - 1 = 9^n.

    But 19 divides 20^m - 1, so 19 divides 9^n, a contradiction. Thus k \ne 1.

    Thus k \in \{ -1, 11 \}.

    Need to show that 20^m = 9^n - 1 has no solutions, and then 11 will be the answer.

    Look mod 10 again, and see that n must be even. Thus need only show that 20^m = 81^n - 1 has no solutions, and then 11 will be the answer.

    Look at 20^m = 81^n - 1 using mod 19: 1 \equiv (5)^n - 1\mod 19, so (5)^n \equiv 2 \mod 19

    Now \mod 19, have: 5^2 \equiv 6, 5^4 \equiv 36 \equiv -2, 5^8 \equiv (-2)^2 \equiv 4, 5^9 \equiv 20 \equiv 1.

    Filling in: 5^3 \equiv 30 \equiv 11, 5^5 \equiv (-2)(5) \equiv -10 \equiv 9, 5^6 \equiv 5^45^2 \equiv -12 \equiv 7,

    and 5^7 \equiv 35 \equiv -3 \equiv 16.

    Therefore (5)^n \equiv 2 \mod 19 has no solutions.

    Therefore 20^m = 81^n - 1 has no solutions.

    Therefore 20^m = 9^n - 1 has no solutions.

    Therefore, k = 11.
    Last edited by johnsomeone; October 5th 2012 at 04:30 PM.
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