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Unformatted text preview: induction: Q(n) = 5n + 1
50 + 1 = 2 = Q(0), which demonstrates that this is true for n = 0.
51 + 1 = 6 = Q(1) (also true for n = 1).
We assume that Q(n1) holds
And that Q(n2) holds
Q(n) = 6(5n1 + 1) – 5(5n2+ 1)
= 6*5n1  5n1 + 6 –5
= 5*5n1 + 1
= 5n + 1. 5n1 + 1
5n2 + 1 3b) Let n = n6n5n4n3n2n1n0 for some decimal number n. Prove that 11n if and only if 11
(n0 – n1 + n2  n3 + n4 – n5 + n6).
n = d0 + 10 * d1 + 100 * d2 + … + 10kdk.
Alternate sum of the digits of number n is: dk – dk1 + dk2  … + (1)kd0
Let us consider the following expression:
E = (10 + 1) * (d0 + 10(d1 – d0) + 100(d2 – d1 – d0) + … + 10k(dk – dk1 + dk2  …
+ (1)kd0)) – 10k+1(dk – dk1 + dk2  … + (1)kd0).
After performing careful multiplication, one can see that E = n.
A term that has a factor of (10 + 1) or 11, is divisible by 11. Thus n is divisible by 11
whenever 10k+1(dk – dk1 + dk2  … + (1)kd0) is divisible by 11. Since 10k+1 is not
divisible by 11 and 11 is a prime, we can make a conclusion that n is divisible by 11 if
and only if dk – dk1 + dk2  … + (1)kd0 is divisible by 11. This number is the alternate
sum of digits of n....
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This document was uploaded on 02/26/2014 for the course ECE 490SIC at Drexel.
 Fall '08
 Trachtenberg

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