GATE | GATE-CS-2005 | Question 13

The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively

(A) 3 and 13
(B) 2 and 11
(C) 4 and 13
(D) 8 and 14


Answer: (C)

Explanation: We know that,
For a number ‘n’, (n) x (n’) = I, where
n’ = inverse of ‘n’
I = Identity element

Now, the identity element for multiplication is 1.
So, we need to find two numbers ‘m’ and ‘n’ such that (4 x m) % 15 = 1 and (7 x n) % 15 = 1,
where ‘m’ is the inverse of 4 and ‘n’ is the inverse of 7, and both ‘m’ and ‘n’ belong to the given set.

Thus, from the given set, it can be easily identified by putting values that m = 4 and n = 13.

So, C is the correct choice.

 
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