SOLUTION: MAT 145: Topics In Contemporary Math
More Modular Arithmetic
Find each of the following.
4) (51 + 17) mod 4
5) (123 + 74) mod 11
6) (302 + 299) mod 7
Algebra ->
Testmodule
-> SOLUTION: MAT 145: Topics In Contemporary Math
More Modular Arithmetic
Find each of the following.
4) (51 + 17) mod 4
5) (123 + 74) mod 11
6) (302 + 299) mod 7
Log On
More Modular Arithmetic
Find each of the following.
4) (51 + 17) mod 4
5) (123 + 74) mod 11
6) (302 + 299) mod 7 Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52781) (Show Source):
(4) Add 51 and 17; then take the remainder of division this sum by 4.
(5) Add 123 and 74; then take the remainder of division this sum by 11.
(6) Add 302 and 299; then take the remainder of division this sum by 7.
You can work this kind of problem mod n by doing the indicated additions and then finding the remainder when the sum is divided by n.
4) (51+17) mod 4 = (68) mod 4; 68/4 = 17 remainder 0 --> answer: 0
5) (123+74) mod 11 = (197) mod 11; 197/11 = 17 remainder 10 --> answer: 10
6) (302+299) mod 7 = (601) mod 7; 601/7 = 85 remainder 6 -- answer: 6
I suspect if this is a learning exercise in modular arithmetic that you were expected to use the features of modular arithmetic, as follows.
4) (51+17) mod 4 = (51) mod 4 + (17) mod 4
51/4 = 12 remainder 3; 17/4 = 4 remainder 1
(51+17) mod 4 = (3+1) mod 4 = (4) mod 4 = 0
5) (123+74) mod 11 = (123) mod 11 + (74) mod 11
123/11 = 11 remainder 2; 74/11 = 6 remainder 8
(123+74) mod 11 = (2+8) mod 11 = 10
6) (302+299) mod 7 = (302) mod 7 + (299) mod 7
302/7 = 43 remainder 1; 299/7 = 42 remainder 5
(302+299) mod 7 = (1+5) mod 7 = 6
In each of these examples, the work is a bit more if you work the problems this second way, because you have to do more "mod" calculations. However, you should know how to work modular arithmetic problems this way because often the work is a lot easier this way.
For example, suppose you had to find (2389041+84932) mod 5. By the first method you would have to add those two large numbers and then find the remainder when the sum is divided by 5. But evaluating a number mod 5 is easy; the quick and easy solution for this example is
(2389041+84932) mod 5 = (2389041) mod 5 + (84932) mod 5 = 1+2 = 3
