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...d1do we can write n = dm · 10" + dm-1 · 10m-1 +...+ do · 10º it follows that: mod (n, 11) = mod (dm · 10m + dm-1 · 10m-1 + ..+ do · 10º, 11) = mod (dm · (-1)™ + dm-1 · (-1)"m-1 + ... = mod ((-1)"(dm – dm-1+ ·· [substitution] + do · (–1)º, 11) [mod(10, 11) = –1] + do · 1), 11) [factor out (-1)"] Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
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...d1do we can write n = dm · 10" + dm-1 · 10m-1 +...+ do · 10º it follows that: mod (n, 11) = mod (dm · 10m + dm-1 · 10m-1 + ..+ do · 10º, 11) = mod (dm · (-1)™ + dm-1 · (-1)"m-1 + ... = mod ((-1)"(dm – dm-1+ ·· [substitution] + do · (–1)º, 11) [mod(10, 11) = –1] + do · 1), 11) [factor out (-1)"] Therefore, mod (n, 11)=0 if and only if the alternating sums of the digits of
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please do Exercise 6.1.22 part abc and please show step by step
![Proposition 6.1.19. An integer is divisible by 3 if and only if the sum of
its digits is divisible by 3.
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...dıdo we can write
n = dm · 10" + dm-1 · 10m-1 + ...+ do · 10º
it follows that:
mod (n, 11)
= mod (dm · 10" + dm-1· 10m–1 +...+ do · 10°, 11)
= mod(dm · (-1)™ + dm-1 · (–1)m-1+.
= mod ((-1)" (dm – dm-1+ ·+do · 1), 11)
[substitution]
..+ do · (–1)º, 11) [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%2F7ecb53a8-32c8-41ee-ae64-6480991310a9%2Ff6aymb3_processed.png&w=3840&q=75)
Transcribed Image Text:Proposition 6.1.19. An integer is divisible by 3 if and only if the sum of
its digits is divisible by 3.
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...dıdo we can write
n = dm · 10" + dm-1 · 10m-1 + ...+ do · 10º
it follows that:
mod (n, 11)
= mod (dm · 10" + dm-1· 10m–1 +...+ do · 10°, 11)
= mod(dm · (-1)™ + dm-1 · (–1)m-1+.
= mod ((-1)" (dm – dm-1+ ·+do · 1), 11)
[substitution]
..+ do · (–1)º, 11) [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.
![Exercise 6.1.22.
(a) In Proposition 6.1.19 we showed that a number is divisible by 3 if and
only if the sum of its digits is divisible by 3. Write a similar argument
and state a proposition for a number that is divisible by 9.
(b) Figure 6.1.1 shows a table giving the different powers of 10 mod base
37.
Based on the results shown in Figure 6.1.1, propose a divisibility rule
to check whether numbers are divisible by 37. Apply your rule to the
following numbers: 17094, 411108, 365412
1 base
37
2
3 10*n(n>o Mod(An, 37)
1
10
10
6.
100
26
7
1000
1
10000
10
100000
26
10
1000000
1
11 10000000
12 100000000
10
26
13
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F7ecb53a8-32c8-41ee-ae64-6480991310a9%2Fdi15eh_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise 6.1.22.
(a) In Proposition 6.1.19 we showed that a number is divisible by 3 if and
only if the sum of its digits is divisible by 3. Write a similar argument
and state a proposition for a number that is divisible by 9.
(b) Figure 6.1.1 shows a table giving the different powers of 10 mod base
37.
Based on the results shown in Figure 6.1.1, propose a divisibility rule
to check whether numbers are divisible by 37. Apply your rule to the
following numbers: 17094, 411108, 365412
1 base
37
2
3 10*n(n>o Mod(An, 37)
1
10
10
6.
100
26
7
1000
1
10000
10
100000
26
10
1000000
1
11 10000000
12 100000000
10
26
13
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.
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