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 Power of two (Posted on 2010-05-12)
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

 Submitted by Ady TZIDON Rating: 4.0000 (1 votes) Solution: (Hide) 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.

 Subject Author Date Proof. Jer 2010-05-12 15:26:52

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