As to Modulus operation I only have seen this form:
(x + y) mod z ≡ K
So I can't understand the question, by the way the answers are :
a) 8 (mod 12) b) 9 (mod 12) c) 10 (mod 12) d) 11 (mod 12) e) None of the above
As to Modulus operation I only have seen this form:
(x + y) mod z ≡ K
So I can't understand the question, by the way the answers are :
a) 8 (mod 12) b) 9 (mod 12) c) 10 (mod 12) d) 11 (mod 12) e) None of the above
The OP may be taking a Computer Science course, in which if $b$ is a positive integer, then $a\bmod{b}$ is the remainder when $a$ is divided by $b$. In that case $\bmod$ is a binary operator. That is different from the $x\equiv y\pmod{m}$ of number theory, which is a ternary relation (or, for fixed $m$, a binary relation).
Calculating $(10+13)\bmod{12}$ is straightforward. Find $10+13$, and calculate the remainder on division by $12$. We get $(10+13)\bmod{12}=11$. So $(10+13)\bmod{12} \equiv 11\pmod{12}$.
Remark: It seems unusual to use the binary operator and the ternary relation in a single short expression.
It is d).
$23 \equiv 11 \text{ mod } 12$ since $(12)(1) + 11 = 23$. In other word, the remainder upon division by $12$ is $11$.