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| Power of two (Posted on 2010-05-12) |
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Given a sequence of of natural numbers
a1, a2, a3, ...
such that a1 is not a multiple of 5
and a(n+1)=a(n)+(the last digit of a(n)).
Prove that this sequence contains infinitely many integer powers of 2.
Ex: 33,36,42,44,48,56,62,
64,....128,.....256 etc
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Submitted by Ady TZIDON
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Rating: 4.0000 (1 votes)
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Solution:
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Let us denote the last digit of a by d.
Since d1 is neither 0 nor 5, d2 is equal to 2, 4, 6, or 8.
Then the sequence d2, d3, ... is periodic with period 4.
Therefore, for each n>1
a(n+4) = a(n )+2+4+6+8=a(n)+20
and a(n+4p)=a(n)+20p for any natural number p.
The sequence {a(n) includes a term divisible by 4, say a(n1) =4k.
But then a(n1+4p) = 4k +20p = 4(k+5p).
Since 2^m (mod5) is a periodic sequence ....1, 2, 4, 3, 1, 2, 4, 3, ..., infinitely many terms 2^m are equal to k mod(5).
Therefore, k+5p contains infinitely many terms of the form 2^m.
Q.E.D. |
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Subject |
Author |
Date |
 | Proof. | Jer | 2010-05-12 15:26:52 |
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