JEE Mathematics 2023 Question with Solution

Given: $x$ and $y$ are distinct integers with $1 \leq x \leq 25$ and $1 \leq y \leq 25$.

Find: The number of ordered choices of $x$ and $y$ such that $x+y$ is divisible by $5$.

Classify the integers from $1$ to $25$ according to their residues modulo $5$. Each residue class has exactly $5$ elements.

For $x+y$ to be divisible by $5$, we need

$$x+y \equiv 0 \pmod{5}$$

So the possible residue pairs are

$$(0,0), (1,4), (2,3), (3,2), (4,1)$$

Now count each case:

• For $(0,0)$, both numbers are multiples of $5$. Since $x$ and $y$ are distinct, the number of ways is

$$\binom{5}{2} \times 2 = 20$$

• For $(1,4)$, the number of ways is

$$5 \times 5 = 25$$

• For $(2,3)$, the number of ways is

$$5 \times 5 = 25$$

• For $(3,2)$, the number of ways is

$$5 \times 5 = 25$$

• For $(4,1)$, the number of ways is

$$5 \times 5 = 25$$

Therefore, the total number of ways is

$$20+25+25+25+25=120$$

Hence, the required number of ways is $120$.

Given: $x$ and $y$ are distinct integers between $1$ and $25$.

Find: The number of choices such that $x+y$ is divisible by $5$.

Using the case split shown in the solution:

• $x=5\lambda, \; y=5\lambda$ gives $20$ ways.
• $x=5\lambda+1, \; y=5\lambda+4$ gives $25$ ways.
• $x=5\lambda+2, \; y=5\lambda+3$ gives $25$ ways.
• $x=5\lambda+3, \; y=5\lambda+2$ gives $25$ ways.
• $x=5\lambda+4, \; y=5\lambda+1$ gives $25$ ways.

Adding all cases,

$$20+25+25+25+25=120$$

Therefore, the total number of ways is $120$.