Key Concepts and Formulas
- Diophantine Equations: Solving linear equations of the form ax+by=c for integer solutions, often with the constraint that the solutions are natural numbers.
- Properties of Complex Numbers: Understanding the cyclic nature of powers of i, where i1=i, i2=−1, i3=−i, and i4=1. This cycle repeats every four powers.
- Factorials and Modular Arithmetic: Understanding that for n≥4, n! is divisible by 4, meaning n!≡0(mod4).
Step-by-Step Solution
Step 1: Determine the number of elements (m) in the set R
The set R is defined as R={(a,b):a+5b=42,a,b∈N}. We need to find the number of pairs (a,b) of natural numbers that satisfy the equation a+5b=42.
Step 2: Express a in terms of b
We rewrite the equation to express a as a function of b:
a=42−5b
Step 3: Apply the natural number constraint for a
Since a∈N, we must have a≥1. Substituting the expression for a, we get:
42−5b≥1
41≥5b
b≤541
b≤8.2
Step 4: Apply the natural number constraint for b
Since b∈N, we must have b≥1.
Step 5: Find all possible integer values for b
Combining the constraints b≥1 and b≤8.2, the possible integer values for b are {1,2,3,4,5,6,7,8}.
Step 6: Calculate the corresponding a values for each b
We iterate through each valid b and find the corresponding a:
- If b=1, a=42−5(1)=37. So, (37,1)∈R.
- If b=2, a=42−5(2)=32. So, (32,2)∈R.
- If b=3, a=42−5(3)=27. So, (27,3)∈R.
- If b=4, a=42−5(4)=22. So, (22,4)∈R.
- If b=5, a=42−5(5)=17. So, (17,5)∈R.
- If b=6, a=42−5(6)=12. So, (12,6)∈R.
- If b=7, a=42−5(7)=7. So, (7,7)∈R.
- If b=8, a=42−5(8)=2. So, (2,8)∈R.
Since all pairs have natural numbers for both a and b, the set R is {(37,1),(32,2),(27,3),(22,4),(17,5),(12,6),(7,7),(2,8)}.
Step 7: Determine m
The number of elements in set R is m=8.
Step 8: Evaluate the summation ∑n=1m(1−in!)
We need to evaluate the sum ∑n=18(1−in!)=x+iy. We can split the sum:
∑n=18(1−in!)=∑n=181−∑n=18in!
Step 9: Calculate ∑n=181
The first part is ∑n=181=1×8=8.
Step 10: Analyze in! for each value of n from 1 to 8
The value of ik depends on k(mod4).
- For n=1: n!=1!=1. i1!=i1=i.
- For n=2: n!=2!=2. i2!=i2=−1.
- For n=3: n!=3!=6. 6(mod4)≡2. i3!=i6=i2=−1.
- For n=4: n!=4!=24. 24(mod4)≡0. i4!=i24=1.
- For n≥4: For any integer n≥4, the term n! will always contain 4 as a factor. This means n! is always a multiple of 4. Therefore, for n=4,5,6,7,8, n!≡0(mod4), which implies in!=1.
Step 11: Calculate the sum ∑n=18in!
∑n=18in!=i1!+i2!+i3!+i4!+i5!+i6!+i7!+i8!
=i+(−1)+(−1)+1+1+1+1+1
=i−2+5
=3+i
Step 12: Calculate the total sum ∑n=18(1−in!)
∑n=18(1−in!)=(∑n=181)−(∑n=18in!)
=8−(3+i)
=8−3−i
=5−i
Step 13: Identify x and y
We have ∑n=1m(1−in!)=x+iy. Comparing 5−i with x+iy, we get x=5 and y=−1.
Step 14: Calculate m+x+y
Substitute the values we found: m=8, x=5, and y=−1.
m+x+y=8+5+(−1)
m+x+y=13−1
m+x+y=12
Common Mistakes & Tips
- Natural Number Definition: Always remember that natural numbers start from 1 (i.e., N={1,2,3,...}).
- Powers of i: Be careful when evaluating powers of i. Remember the cycle: i1=i,i2=−1,i3=−i,i4=1. Also, remember to use modulo 4.
Summary
By carefully analyzing the Diophantine equation and the properties of complex numbers, we found m=8, x=5, and y=−1. Therefore, m+x+y=12.
The final answer is \boxed{12}, which corresponds to option (A).