Re: irrational numbers
- From: Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx>
- Date: 25 Apr 2007 06:13:12 -0400
quasi <quasi@xxxxxxxx> wrote:
shree_arpus12389@xxxxxxxxx wrote:
how to find the units digit in the exression:
(15+sqrt220)^19?
The units digit of (15+sqrt(220))^n is 9 for all integers n>=1.
That is true for all j + sqrt(k) with (j-1)^2 < k < j^2, 5|j,k
as the proof below shows. For a deeper understanding of this
one may study Lucas sequences and quadratic number fields.
--Bill Dubuque
It can be explained as follows ....
Let u1=15+sqrt(220) and u2=15-sqrt(220).
Note that 0 < u2 < 1.
Then u1 and u2 are roots of the quadratic equation
(0) u^2=30*u-5
For the rest of the discussion, the variable n will be assumed to
represent a nonnegative integer.
Then equation (0) implies
(1) u1^(n+2) = 30*u1^(n+1) - 5*u1^n
(2) u2^(n+2) = 30*u2^(n+1) - 5*u2^n
for all n.
Let w[n]=u1^n+u2^n
Since u1 and u2 are positive, w[n] is positive for all n.
Adding equations (1) and (2) gives
(3) w[n+2] = 30*w[n+1] - 5*w[n] for all n.
By direct calculation, w[0]=2 and w[1]=30.
Since w[0] and w[1] are integers, w[n] is an integer for all n, hence
w[n] is a positive integer for all n.
From equation (3), w[2]=890.
Sincce w[1] and w[2] are multiples of 10, w[n] is a multiple of 10 for
all n>=1.
Hence, for all n>=1, w[n] is a positive integer with units digit 0.
But u1^n = w^n - u2^n, and 0 < u2^n < 1 for all n>=1, so
w[n] - 1 < u1^n < w[n]
It follows that the units digit of w[n] is 9 for all n>=1.
quasi
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