Find the smallest multiple of 9 with no odd digits.

Question

Homi · Accepted Answer

We need to find a number where all its digits are even, and the sum of its digits is a multiple of 9.

Step 1 — Understand the conditions

First, let us understand what the question asks for. A multiple of 9 is a number that can be divided by 9 without a remainder. The divisibility rule for 9 states that a number is a multiple of 9 if the sum of its digits is a multiple of 9. "No odd digits" means that every digit in the number must be an even digit. The even digits are 0, 2, 4, 6, 8.

Step 2 — Find the smallest number of digits

To find the smallest multiple, we should start by looking for numbers with the fewest possible digits.

Let us check 1-digit numbers. The 1-digit even numbers are 0, 2, 4, 6, 8. None of these (except 0) are multiples of 9. We are looking for a positive multiple of 9.

Let us check 2-digit numbers. Let the number be $AB$ , where $A$ and $B$ are even digits. The sum of the digits, $A+B$ , must be a multiple of 9. The smallest possible sum of two even digits (where $A \neq 0$ ) is $2+0=2$ . The largest possible sum of two even digits is $8+8=16$ . So, the sum of digits $A+B$ must be 9. However, if $A$ and $B$ are both even, their sum $A+B$ must also be an even number. Since 9 is an odd number, it is not possible to get a sum of 9 from two even digits. Therefore, there are no 2-digit multiples of 9 with no odd digits.

Let us check 3-digit numbers. Let the number be $ABC$ , where $A, B, C$ are even digits. The first digit $A$ cannot be 0. So $A \in \{2, 4, 6, 8\}$ . The sum of the digits, $A+B+C$ , must be a multiple of 9. The smallest possible sum of three even digits (where $A \neq 0$ ) is $2+0+0=2$ . The largest possible sum of three even digits is $8+8+8=24$ . So, the sum of digits $A+B+C$ must be either 9 or 18.

Can the sum be 9? If $A, B, C$ are all even, their sum $A+B+C$ must be an even number. Since 9 is an odd number, it is not possible to get a sum of 9 from three even digits. Therefore, the sum of the digits must be 18.

Step 3 — Construct the smallest number

We need a 3-digit number $ABC$ where $A, B, C$ are even digits, $A \neq 0$ , and $A+B+C=18$ . To make the number as small as possible, we want:

The smallest possible first digit ( $A$ ).
The smallest possible second digit ( $B$ ).
The smallest possible third digit ( $C$ ).

Let us try the smallest possible value for $A$ , which is 2. If $A=2$ , then $B+C = 18 - 2 = 16$ . Now we need to find two even digits $B$ and $C$ that add up to 16. To make the number $2BC$ smallest, we want $B$ to be as small as possible. Possible even digits for $B$ are $0, 2, 4, 6, 8$ . If $B=0$ , then $C=16$ . But 16 is not a single digit. If $B=2$ , then $C=14$ . Not a single digit. If $B=4$ , then $C=12$ . Not a single digit. If $B=6$ , then $C=10$ . Not a single digit. If $B=8$ , then $C=8$ . This works! Both 8 and 8 are even digits. So, if $A=2$ , the smallest number we can form is 288. Let us check: $2+8+8 = 18$ . 18 is a multiple of 9. All digits are even.

Let us check if starting with a larger $A$ would give a smaller number. If $A=4$ , then $B+C = 18 - 4 = 14$ . To make $4BC$ smallest, we want $B$ to be as small as possible. Possible pairs for $(B, C)$ that sum to 14: If $B=6$ , then $C=8$ . This gives the number 468. If $B=8$ , then $C=6$ . This gives the number 486. Both 468 and 486 are larger than 288.

Any number starting with $A=6$ or $A=8$ will be even larger. For example, if $A=6$ , then $B+C = 18 - 6 = 12$ . The smallest $B$ could be 4 (since $C$ would be 8). This gives 648. This is also larger than 288.

So, the smallest multiple of 9 with no odd digits is 288.

$\boxed{288}$

Answer

(i) The smallest multiple of 9 with no odd digits is 288.

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