JEE 2023 — Mathematics PYQ

We categorize the integers from $1$ to $25$ based on their remainder when divided by $5$:

• $R_0=\{5,10,15,20,25\}$ (Numbers of the form $5k$) — 5 numbers

• $R_1=\{1,6,11,16,21\}$ (Numbers of the form $5k+1$) — 5 numbers

• $R_2=\{2,7,12,17,22\}$ (Numbers of the form $5k+2$) — 5 numbers

• $R_3=\{3,8,13,18,23\}$ (Numbers of the form $5k+3$) — 5 numbers

• $R_4=\{4,9,14,19,24\}$ (Numbers of the form $5k+4$) — 5 numbers

Step 2: Conditions for $x+y$ to be divisible by 5

For the sum $x+y$ to be divisible by $5$, the sum of their remainders must be $0$ or $5$. Since $x$ and $y$ must be distinct ($x\ne y$), we consider the following cases:

1. Case 1: Both $x$ and $y$ are from $R_0$

   Both numbers leave a remainder of $0$. We need to choose $2$ distinct numbers from the $5$ available in $R_0$.

   $$\text{Number of ways}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5\times 4}{2}=10$$

2. Case 2: One number from $R_1$ and one from $R_4$

   The remainders are $1$ and $4$, so $1+4=5$ (divisible by $5$).

   $$\text{Number of ways}=5\times 5=25$$

3. Case 3: One number from $R_2$ and one from $R_3$

   The remainders are $2$ and $3$, so $2+3=5$ (divisible by $5$).

   $$\text{Number of ways}=5\times 5=25$$

Step 3: Total number of ways

If we are choosing an unordered pair $\{x,y\}$:

If the question implies ordered pairs $(x,y)$ (where the order of selection matters):