Transcript Modular Arithmetic - UTEP Math Department

Lecture 12
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The ISBN 10-digit uses a reverse weighting
system: multiply the first digit by 10, the second
by 9, the third by 8 and so on until the check
digit (the 10th digit) is multiplied by 1. The sum
of these should be evenly divisible by 11.
The ISBN 13-digit uses a different system: add all
the digits in the odd places values; add all the
digits in the even place values and multiply this
sum by 3; finally add the two results. This result
should be evenly divisible by 10.
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A certain textbook has ISBN 1-285-56352-C,
where C is the check digit. Let’s determine
what that number should be.
First we multiply appropriately:
10(1)  9  2   8  8   7  5   6  5   5  6   4  3   3  5   2  2   1  C 
 10  18  64  35  30  30  12  15  4  C
 218  C
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We now need 218 + C to be divisible by 11.
Multiples of 11 (near where we care) are 198,
209, and 220.
If we add 218 + 2 we will get an even
multiple of 11, namely 220. Therefore C = 2
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Using the 10-digit ISBN 1-285-56352-2 from our
previous example, we will now convert it to a 13-digit
ISBN.
First, attach 978 to the front of the number and
remove the original check digit in order to get 9781-285-56352-C, where C is the new check digit.
Next, add all the digits in the odd place values:
9+8+2+5+6+5+C = 35 + C.
Then add all the digits in the even place values and
multiply the result by 3: 3(7+1+8+5+3+2) = 3(26) =
78.
Add these to get 35 + C + 78 = 113 + C. We need
this to be divisible by 10 so C = 7.
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To find the check digit in a UPC code we
multiply the digits in the odd place values by
3 and the digits in the even place values by 1.
Add these results to a value that is divisible
by 10.
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Find the missing check digit in the product
with UPC 0 44700 07205 C.
Multiply the digits in odd place values by 3 to
get: 3(0+4+0+0+2+5) = 3(11) = 33.
Multiply the digits in the even place values by
1 to get: 1(4+7+0+7+0+C) = 18 + C.
Add these together to get 33 + 18 + C = 51
+ C.
It must be that C = 9 so that the result is a
multiple of 10.
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A correct bank routing number is determined
by multiplying the digits of the routing
number by 7, 3, 9, 7, 3, 9, 7, 3, 9, adding the
results and the final sum must be divisible by
10.
Bank routing numbers do not have a
traditional check digit, it is more of an error
detecting code in order to verify the number
given is correct.
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I think my Wells Fargo routing number is
107002129. Let’s check:
7(1)+3(0)+9(7)+7(0)+3(0)+9(2)+7(1)+3(2)+9
(9) = 7+63+18+7+6+81 = 182.
This number must be incorrect because 182
is not evenly divisible by 10.
In order to determine the correct error,
probably due to a transposition, I will switch
pairs of numbers until I find the correct code.
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To determine the last digit of a 16-digit
credit card number:
◦ Add the digits in the odd place values and double
the result.
◦ Next, count the number of digits in the odd place
values that are greater than 4 and add this to the
total.
◦ Finally, add the previous result to the sum of the
digits in the even place values.
◦ The final sum must be divisible by 10.
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A card is issued with the number 3125 6001
9643 001C, where C is the check digit. Let’s find
it.
Add digits in the odd place values and double the
result: 2(3+2+6+0+9+4+0+1) = 2(25) = 50.
Count the number of digits in odd place values
that are more than 4: only 2 (the 6 and 9) so we
add 2 + 50 to get 52.
Add the values in even place values to 52 to get:
52+1+5+0+1+6+3+0+C = 68 + C.
In order for this to be divisible by 10 C must be
2.
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Add the first 10 digits and divide by 9. The
check digit is the remainder.
Example: A money order has number
6312472907C. Find the check digit.
◦ Add the first ten digits to get
6+3+1+2+4+7+2+9+0+7 = 41.
◦ Divide 41 by 9 to get 41 = 9(4) + 5, where 5 is the
remainder.
◦ It must be that C = 5.