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

Understand the Question

A number is divisible by $15$ if it is divisible by both $3$ (sum of digits is a multiple of $3$ ) and $5$ (last digit is $0$ or $5$ ).
A number is divisible by $6$ if it is divisible by both $2$ (last digit is even) and $3$ .
For a number to remain a $6$ -digit number when reversed, neither its first digit nor its last digit can be $0$ .

Step 1 · Determine the First and Last Digits

Let the $6$ -digit number be $N = d_5 d_4 d_3 d_2 d_1 d_0$ and the reversed number be $N_R = d_0 d_1 d_2 d_3 d_4 d_5$ .

Finding $d_0$ :
- $N$ is divisible by $15 \implies N$ is divisible by $5 \implies d_0 \in \{0, 5\}$ .
- Since $N_R$ must also be a $6$ -digit number, the leading digit $d_0 \neq 0$ .
- Therefore, $d_0 = 5$ .
Finding $d_5$ :
- $N_R$ is divisible by $6 \implies N_R$ is divisible by $2 \implies$ last digit $d_5$ is even.
- Since $N$ is a $6$ -digit number, $d_5 \neq 0$ .
- Therefore, $d_5 \in \{2, 4, 6, 8\}$ .

Step 2 · Construct the Number

Both $N$ and $N_R$ must be divisible by $3$ , which means the sum of digits $S$ must be a multiple of $3$ .

Let $d_5 = 2$ and $d_0 = 5$ . Set the middle digits $d_1 = 0, d_2 = 0, d_3 = 0$ .

\begin{aligned} S &= d_5 + d_4 + d_3 + d_2 + d_1 + d_0 \\[0.6em] &= 2 + d_4 + 0 + 0 + 0 + 5 \\[0.6em] &= 7 + d_4 \end{aligned}

For $S$ to be divisible by $3$ , choose $d_4 = 2$ (so $S = 9$ ).

Thus, the $6$ -digit number is $220005$ .

Step 3 · Verify the Solution

Check $220005$ for divisibility by $15$ :
- Last digit is $5 \implies$ divisible by $5$ .
- Sum of digits $= 2 + 2 + 0 + 0 + 0 + 5 = 9$ (divisible by $3$ ) $\implies$ divisible by $3$ .
- Therefore, $220005$ is divisible by $15$ .
Check reversed number $500022$ for divisibility by $6$ :
- Last digit is $2$ (even) $\implies$ divisible by $2$ .
- Sum of digits $= 5 + 0 + 0 + 0 + 2 + 2 = 9$ (divisible by $3$ ) $\implies$ divisible by $3$ .
- Therefore, $500022$ is divisible by $6$ .

Answer

$220005$

Common Mistakes

Choosing $d_0 = 0$ : Although a number ending in $0$ is divisible by $5$ , reversing it would give a leading zero, reducing it to a $5$ -digit number.
Ignoring the Divisibility by 3 Rule: Divisibility by $15$ and $6$ both require the sum of digits to be a multiple of $3$ .
Non-Unique Answers: Multiple valid $6$ -digit numbers exist (e.g., $200025, 410005$ ); ensure the constructed number satisfies all divisibility and digit count criteria.

Question 26