Create a spreadsheet similar to the the spreadsheet in Figure 6.1.1. Use your spreadsheet to find mod (10", 111) for 0 ≤ n ≤ 8. Come up with a proposition for numbers in base 111 and prove it similarly the divisibility rule for numbers in base 11 was proved in Proposition 6.1.21.
Create a spreadsheet similar to the the spreadsheet in Figure 6.1.1. Use your spreadsheet to find mod (10", 111) for 0 ≤ n ≤ 8. Come up with a proposition for numbers in base 111 and prove it similarly the divisibility rule for numbers in base 11 was proved in Proposition 6.1.21.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please do part C and just do the proposition part you do not have to worry about the excel part. Thanks
Transcribed Image Text:1
2
3
4
5
6
7
8
A
base
10^n(n>=0) Mod(An,37)
1
10
100
1000
10000
9
10
11 10000000
12 100000000
13
100000
1000000
37
1
10
26
1
10
26
1
10
26
с
Figure 6.1.1. Spreadsheet to compute the powers of 10 mod 37
(c) Create a spreadsheet similar to the the spreadsheet in Figure 6.1.1. Use
your spreadsheet to find mod (10", 111) for 0 ≤ n ≤ 8. Come up with a
proposition for numbers in base 111 and prove it similarly the divisibility
rule for numbers in base 11 was proved in Proposition 6.1.21.
![with -1 whenever we are taking mod's base 10.
mod (6472, 11) = mod (6 - 10³ +4·10² +7-10+2 -1,11)
mod (6 (-1)³ +4 (−1)² +7- (-1)+2, 11)
= mod(−6+4−7+2, 11) = mod(-7, 11) = 4
Since mod (6472, 11) #0, 6472 is not divisible by 11.
Proposition 6.1.21. A number is divisible by 11 if and only if the alternat-
ing sums of the digits is divisible by 11. (Note: alternating sums is where
the signs of the number alternate when summing.)
PROOF. Given an integer with digits do...dn where the number is writeen
as dndn-1...dido we can write
n=dm 10" + dm-1-10-1
it follows that:
++ do 10⁰
mod (n, 11)
= mod (dm - 10 + dm-1·10m-¹ +... + do 10º, 11)
= mod (dm - (-1) + dm-1 · (−1)m-1 +...+do
= mod ((-1) (dm - dm-1 ++ do 1), 11)
(-1)0,11)
[substitution]
[mod (10, 11) = -1]
[factor out (-1)]
Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
the number d...do is divisible by 11.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F9677c610-58c9-45a6-951b-0ca39f8544cb%2F2xto5he_processed.png&w=3840&q=75)
Transcribed Image Text:with -1 whenever we are taking mod's base 10.
mod (6472, 11) = mod (6 - 10³ +4·10² +7-10+2 -1,11)
mod (6 (-1)³ +4 (−1)² +7- (-1)+2, 11)
= mod(−6+4−7+2, 11) = mod(-7, 11) = 4
Since mod (6472, 11) #0, 6472 is not divisible by 11.
Proposition 6.1.21. A number is divisible by 11 if and only if the alternat-
ing sums of the digits is divisible by 11. (Note: alternating sums is where
the signs of the number alternate when summing.)
PROOF. Given an integer with digits do...dn where the number is writeen
as dndn-1...dido we can write
n=dm 10" + dm-1-10-1
it follows that:
++ do 10⁰
mod (n, 11)
= mod (dm - 10 + dm-1·10m-¹ +... + do 10º, 11)
= mod (dm - (-1) + dm-1 · (−1)m-1 +...+do
= mod ((-1) (dm - dm-1 ++ do 1), 11)
(-1)0,11)
[substitution]
[mod (10, 11) = -1]
[factor out (-1)]
Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
the number d...do is divisible by 11.
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