1) Show that a positive integer m is divisible by 11 if and only if the al- tern
ID: 3199282 • Letter: 1
Question
1) Show that a positive integer m is divisible by 11 if and only if the al- ternating sum of its digits is divisible by 11. Hint: notice that 10-1 (mod 1) 2) Is 2121212121212121212121 divisible by 9? Is it divisible by 11? 3) The number 3-540-97285-9 is obtained from a valid ISBN number by switching two consccutive digits. Find the ISBN number. 4) The number 0-31-03()369-0 is obtained fron a valid ISBN number by switching two consecutive digits. Find the ISBN number 5) Let n be a positive integer. If y (mod n) and yz (mod n) prove that z (mod n) for any three integers z, y and zExplanation / Answer
1. We know that
10 = -1(mod 11)
100 = 10 x 10 = (-1)(mod 11) x (-1)(mod 11) = (-1) x (-1) (mod 11) = 1 (mod 11)
Similarly 1000 = -1(mod 11)
Thus in general 10n = (-1)n (mod 11)
A number in decimal system can be written as
N = 10n xn + 10 (n-1) xn-1 + 10(n-2) x n-2 + . …….+102 x2 +101 x1 +100 xo
So when N is divided by 11, then N(mod 11) = 0
Therefore R H S (mod 11) should also be = 0
R HS mod(11) is the sum of all the (mod 11) taken together
Thus R HS (mod 11) is = 10n xn + 10 (n-1) xn-1 + 10(n-2) x n-2 + . …….+102 x2 +101 x1 +100 xo) (mod 11)
100 xo (mod 11) = xo
101 x1 (mod 11) = - x1
102 x2 (mod 11) = x2
……
Total therefore is the alterating sum of its digits. This should be zero.
The number N is divisible by eleven if and only if the alternating sum of its digits is divisible by 11.
Example:
1547647
We should calculate +7-4+6-7+4-5+1 = +2. Not equal to sero
Hence not divisible by 11
And the reminder is 2 when divided by 11
2.
2121212121212121212121
+1-2+1-2+1-2+1-2+1-2+1-2+1-2+1-2+1-2+1-2+1-2 = -11
Mod 11 of this is zero.
Hence the number is divisibie by 11
The divisibility test for 9 or Mod 9 is the sum of thew digits should be divisible by 9
2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1 = 33
33 is not divisible by 9. Hence 2121212121212121212121 not divisbly by 9
3. ISBN Checksum is obtained by multiplying digits from the left side by successively by 10, 9, 8, 7, 6, 5, 4, 3 and 2. And then adding these numbers which is divided by 11. The reminder is the check sum.
So to get the original ISBN calculate the same way the check sum and find how the difference3 has taken place.
3
10
30
5
10
50
5
9
45
3
9
27
4
8
32
4
8
32
0
7
0
0
7
0
9
6
54
9
6
54
7
5
35
7
5
35
2
4
8
2
4
8
8
3
24
8
3
24
5
2
10
5
2
10
9
9
sum
238
240
Mod (11)
7
Mod (11)
9
The original ISBN number is 5340972859
4. Given number 0310303690
Calculation is as follows
0
10
0
0
10
0
3
9
27
1
9
9
1
8
8
3
8
24
0
7
0
0
7
0
3
6
18
3
6
18
0
5
0
0
5
0
3
4
12
3
4
12
6
3
18
6
3
18
9
2
18
9
2
18
0
0
sum
101
99
Mod (11)
2
Mod (11)
0
The ISBN Number is 0130303690
5. By rule of modules multiplication , if x, y and z are integers and n is a positive integer such that
x = y(mod n) and y = z(mod n)
then by multiplying both sides we get x .y = y.z (mod n)
Cancelling common factor y, we get x = z(mod n)
3
10
30
5
10
50
5
9
45
3
9
27
4
8
32
4
8
32
0
7
0
0
7
0
9
6
54
9
6
54
7
5
35
7
5
35
2
4
8
2
4
8
8
3
24
8
3
24
5
2
10
5
2
10
9
9
sum
238
240
Mod (11)
7
Mod (11)
9
the numbers are 3 and 5 reversedRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.