Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Question

Homi · Accepted Answer

We need to find a 6-digit number that satisfies two conditions using divisibility rules.

Step 1 — Understanding Divisibility Rules

A number is divisible by 15 if it is divisible by both 3 and 5. A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 5 if its last digit is 0 or 5.

A number is divisible by 6 if it is divisible by both 2 and 3. A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, 8). A number is divisible by 3 if the sum of its digits is divisible by 3.

Step 2 — Determining the First and Last Digits

Let the 6-digit number be $N$ . We can write its digits as $d_5 d_4 d_3 d_2 d_1 d_0$ . Here, $d_5$ is the first digit and $d_0$ is the last digit. Since $N$ is a 6-digit number, its first digit $d_5$ cannot be 0.

The number $N$ is divisible by 15. This means $N$ must be divisible by 5. So, its last digit, $d_0$ , must be 0 or 5.

Now, let's consider the reversed number, $N_R$ . The digits of $N_R$ are $d_0 d_1 d_2 d_3 d_4 d_5$ . Since $N_R$ is also a 6-digit number, its first digit, $d_0$ , cannot be 0. This means $d_0$ must be 5. So, the last digit of our original number $N$ is 5.

The reversed number $N_R$ is divisible by 6. This means $N_R$ must be divisible by 2. So, its last digit, $d_5$ , must be an even number. Since $d_5$ cannot be 0 (as $N$ is a 6-digit number), $d_5$ must be 2, 4, 6, or 8.

$\boxed{d_0 = 5 \text{ and } d_5 \in \{2, 4, 6, 8\}}$

Step 3 — Satisfying Divisibility by 3

Both the original number $N$ and the reversed number $N_R$ must be divisible by 3. The sum of the digits of $N$ is $S = d_5 + d_4 + d_3 + d_2 + d_1 + d_0$ . The sum of the digits of $N_R$ is also $S$ . For a number to be divisible by 3, its sum of digits must be divisible by 3. So, $S$ must be a multiple of 3.

Step 4 — Constructing the Number

We know $d_0 = 5$ . We can choose any digit from $\{2, 4, 6, 8\}$ for $d_5$ . Let us choose the smallest, $d_5 = 2$ . So our number looks like $2 d_4 d_3 d_2 d_1 5$ .

The sum of its digits is $S = 2 + d_4 + d_3 + d_2 + d_1 + 5$ . $S = 7 + d_4 + d_3 + d_2 + d_1$ We need $S$ to be a multiple of 3. To find a simple number, let us choose the middle digits $d_1, d_2, d_3$ to be 0. So, $d_1 = 0$ , $d_2 = 0$ , $d_3 = 0$ . Now the sum becomes $S = 7 + d_4 + 0 + 0 + 0$ . $S = 7 + d_4$ We need $7 + d_4$ to be a multiple of 3. Let us try values for $d_4$ : If $d_4 = 0$ , $S = 7$ , not divisible by 3. If $d_4 = 1$ , $S = 8$ , not divisible by 3. If $d_4 = 2$ , $S = 9$ , which is divisible by 3. So, we can choose $d_4 = 2$ .

The digits are $d_5=2, d_4=2, d_3=0, d_2=0, d_1=0, d_0=5$ . The 6-digit number is 220005.

Step 5 — Verifying the Solution

Let us check if the number 220005 meets all conditions.

Is 220005 divisible by 15?
- Divisibility by 5: The last digit is 5. Yes, it is divisible by 5.
- Divisibility by 3: The sum of digits is $2+2+0+0+0+5 = 9$ . Since 9 is divisible by 3, the number is divisible by 3.
- Since it is divisible by both 3 and 5, it is divisible by 15.
When the digits are reversed, is the new number divisible by 6?
- The reversed number is 500022.
- Divisibility by 2: The last digit is 2 (an even number). Yes, it is divisible by 2.
- Divisibility by 3: The sum of digits is $5+0+0+0+2+2 = 9$ . Since 9 is divisible by 3, the number is divisible by 3.
- Since it is divisible by both 2 and 3, it is divisible by 6.

Both conditions are satisfied.

Answer

(i) The 6-digit number is 220005.

Question 26

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