Correct answers only !
what is the check digit, d, for the visa number shown below?
4328 3103 2635 108d
a. 1
b. 3
c. 5
d. 7

D.) 7

Step-by-step explanation:

To calculate the check digit, multiply every even-position digit (when counted from the right) in the number by two. If the result is a two digit number, then add these digits together to make a single digit (this is called the digital root):

Odd numbers:  (1.)= 8 (3.)= 4 (5.)= 6 (7.)= 0 (9.)= 4 (11.)= 6 (13.)= 2 (15.)=(1 + 6)= 7

Even numbers: (2.)= 3 (4.)= 8 (6.)= 1 (8.)= 3 (10.)= 6 (12.)=5 (14.)=0

To this total, we then add every odd-position digit:

Odd = 37

Even = 26

This will result in a total:

37 + 26 = 63

The check-digit is what number needs to be added to this total to make the next multiple of 10:

Next multiple of 10 is 70.

70 - 63 = 7

7

Hope this helps : D

  D.  7

Step-by-step explanation:

Divide the number into odd-numbered and even-numbered digits (starting counting digit number 1 from the right).

  odd digits: d, 0, 5, 6, 3, 1, 8, 3 . . . . sum of 26+d

  even digits: 8, 1, 3, 2, 0, 3, 2, 4 . . . sum of 23

Multiply the sum of even-numbered digits by 2, and add 1 for each even-numbered digit* that is 5 or greater: 23×2 +1 = 47

Add this to the sum of odd-numbered digits, and choose "d" so the result is a multiple of 10.

  47 + (26+d) = 73+d

The next higher multiple of 10 is 80. For the total to be 80, d = 7.

* The actual algorithm is described as "for each product of a digit and 2 that is greater than 9, subtract 9 from the product." Doing that, the number added to the sum will be ...

In each case, the number contributed to the sum is 1 more than twice the digit, mod 10.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!