JEE 2022 — Mathematics PYQ

Solution

Let the four-digit number be represented as $d_1{d}_2{d}_3{d}_4$.

According to the problem:

• $d_1\in \{1,2,\dots ,9\}$ (The first digit cannot be zero).

• $d_2,d_3\in \{0,1,\dots ,9\}$.

• $d_4\in \{1,2,\dots ,9\}$ (The last digit cannot be zero because we cannot divide by zero).

• $d_1,d_2,$ and $d_3$ must all be divisible by $d_4$.

We will calculate the number of possibilities for each case of $d_4$:

Case 1: $d_4=1$

Every digit is divisible by $1$.

• $d_1$: 9 options ($1$ to $9$)

• $d_2$: 10 options ($0$ to $9$)

• $d_3$: 10 options ($0$ to $9$)

• Total $=9\times 10\times 10=900$

Case 2: $d_4=2$

The digits must be multiples of $2$ (specifically $0,2,4,6,8$).

• $d_1$: 4 options ($2,4,6,8$)

• $d_2$: 5 options ($0,2,4,6,8$)

• $d_3$: 5 options ($0,2,4,6,8$)

• Total $=4\times 5\times 5=100$

Case 3: $d_4=3$

The digits must be multiples of $3$ ($0,3,6,9$).

• $d_1$: 3 options ($3,6,9$)

• $d_2$: 4 options ($0,3,6,9$)

• $d_3$: 4 options ($0,3,6,9$)

• Total $=3\times 4\times 4=48$

Case 4: $d_4=4$

The digits must be multiples of $4$ ($0,4,8$).

• $d_1$: 2 options ($4,8$)

• $d_2,d_3$: 3 options each ($0,4,8$)

• Total $=2\times 3\times 3=18$

Case 5: $d_4=5$

The digits must be multiples of $5$ ($0,5$).

• $d_1$: 1 option ($5$)

• $d_2,d_3$: 2 options each ($0,5$)

• Total $=1\times 2\times 2=4$

Case 6 to 9: $d_4\in \{6,7,8,9\}$

For these cases, only $0$ and the digit itself are multiples.

• $d_1$: 1 option (the digit itself)

• $d_2,d_3$: 2 options each ($0$ and the digit)

• Total per case $=1\times 2\times 2=4$

• For 4 cases ($6,7,8,9$): $4\times 4=16$

Final Calculation

Summing all the cases together:

Final Answer:

The total number of such four-digit numbers is 1086.