Step 1: Understand the operation.
The operation ∗ is defined as a∗b=(ab+3)(mod8). This means we multiply a and b, add 3, and then find the remainder when divided by 8. The set is P={1,3,5,6}.
Step 2: Calculate the missing values for the table.
We will calculate a∗b for each pair (a,b) from the set P.
For a=1:
1∗1=(1×1+3)(mod8)=(1+3)(mod8)=4(mod8)=4
1∗3=(1×3+3)(mod8)=(3+3)(mod8)=6(mod8)=6
1∗5=(1×5+3)(mod8)=(5+3)(mod8)=8(mod8)=0
1∗6=(1×6+3)(mod8)=(6+3)(mod8)=9(mod8)=1
For a=3:
3∗1=(3×1+3)(mod8)=(3+3)(mod8)=6(mod8)=6
3∗3=(3×3+3)(mod8)=(9+3)(mod8)=12(mod8)=4
3∗5=(3×5+3)(mod8)=(15+3)(mod8)=18(mod8)=2
3∗6=(3×6+3)(mod8)=(18+3)(mod8)=21(mod8)=5
For a=5:
5∗1=(5×1+3)(mod8)=(5+3)(mod8)=8(mod8)=0
5∗3=(5×3+3)(mod8)=(15+3)(mod8)=18(mod8)=2
5∗5=(5×5+3)(mod8)=(25+3)(mod8)=28(mod8)=4
5∗6=(5×6+3)(mod8)=(30+3)(mod8)=33(mod8)=1
For a=6:
6∗1=(6×1+3)(mod8)=(6+3)(mod8)=9(mod8)=1
6∗3=(6×3+3)(mod8)=(18+3)(mod8)=21(mod8)=5
6∗5=(6×5+3)(mod8)=(30+3)(mod8)=33(mod8)=1
6∗6=(6×6+3)(mod8)=(36+3)(mod8)=39(mod8)=7
Step 3: Complete the table with the calculated values.
The completed table is:
∗135614601364255024161517
The completed table is:
\begin{array{|c|c|c|c|c|}
\hline
* & 1 & 3 & 5 & 6 \\
\hline
1 & 4 & 6 & 0 & 1 \\
\hline
3 & 6 & 4 & 2 & 5 \\
\hline
5 & 0 & 2 & 4 & 1 \\
\hline
6 & 1 & 5 & 1 & 7 \\
\hline
\end{array}
}
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