Question

Suppose \(A\) is denoted the set of all numbers between 1 and 700 which are divisible by 3 and let \(B\) is denoted the set of all numbers between 1 and 300 which are divisible by 7. If \(C=\{(a,b)\mid a\in A,b\in B, a\ne b \text{ and } a+b=\text{even number}\}\), then order of \(C\) is:

• A. 4879
• B. 4789
• C. 6789
• D. 9876

Hint:

Always double-check for subtraction constraints like $a \neq b$. Finding common multiples of the divisors ($\text{LCM}(3, 7) = 21$) inside the lower boundary domain is the fastest way to pull out the duplicate elements!

Answer 1

The Correct Option is A

Concept: The number of coordinate pairs $(a,b)$ yielding an even sum constraint can be found by evaluating set cardinalities across odd and even sub-parities. For the sum $(a+b)$ to be an even integer, both numbers must share identical parity properties: $$\text{Even} + \text{Even} = \text{Even} \quad \text{and} \quad \text{Odd} + \text{Odd} = \text{Even}$$ Step 1: Determine the elements and parity split of set A.
Set $A$ contains elements between 1 and 700 divisible by 3: $A = \{3, 6, 9, \dots, 699\}$. The total number of terms is $N_A = \lfloor \frac{699}{3} \rfloor = 233$. Since the sequence alternates uniformly between odd and even integers, starting and ending on odd numbers, there is exactly one extra odd integer: $$n(A_{\text{odd}}) = \frac{233 + 1}{2} = 117 \quad \text{and} \quad n(A_{\text{even}}) = 233 - 117 = 116$$

Step 2: Determine the elements and parity split of set B.
Set $B$ contains elements between 1 and 300 divisible by 7: $B = \{7, 14, 21, \dots, 294\}$. The total number of terms is $N_B = \lfloor \frac{294}{7} \rfloor = 42$. Since 42 is an even integer, the set splits perfectly into equal components of odd and even numbers: $$n(B_{\text{odd}}) = \frac{42}{2} = 21 \quad \text{and} \quad n(B_{\text{even}}) = \frac{42}{2} = 21$$

Step 3: Calculate base combinations matching the even sum condition.
Using the fundamental multiplication principle across the matching parity sets:
• Both elements are odd: $\text{Ways}_1 = n(A_{\text{odd}}) \times n(B_{\text{odd}}) = 117 \times 21 = 2457$
• Both elements are even: $\text{Ways}_2 = n(A_{\text{even}}) \times n(B_{\text{even}}) = 116 \times 21 = 2436$ Summing these separate configurations gives the base count: $$\text{Total Parity Pairs} = 2457 + 2436 = 4893$$

Step 4: Subtract the overlapping elements where $a = b$.
The question specifies that the pairs must satisfy $a \neq b$. We must find how many identical values are shared by both sets up to the lower domain limit of 300. Shared values are common multiples of 3 and 7, which means multiples of 21: $$\text{Shared elements} = \lfloor \frac{294}{21} \rfloor = 14 \text{ elements}$$ Since any number added to itself automatically satisfies the even constraint ($a + a = 2a$), all 14 of these matching duplicate elements were counted in our Step 3 total. Subtracting these invalid overlapping elements: $$\text{Order of C} = 4893 - 14 = 4879$$ This matches option (A).