The two-dight number that is divisible by both the sum and the product of its dights?
Well, it cant be greater than 81. So work down a common multiply table..
I found it within one minute.
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To find a two-digit number that is divisible by both the sum and the product of its digits, we need to follow these steps:
1. Let's consider all the possible two-digit numbers from 10 to 99.
2. For each number, we will calculate the sum and the product of its digits.
3. We'll check if the number is divisible by both the sum and the product.
4. If we find a number that meets both conditions, we can stop and consider it as the answer.
Let's perform these steps together:
Starting with the number 10:
- The sum of its digits is 1 + 0 = 1, and the product is 1 × 0 = 0. However, 10 is not divisible by 1 or 0, so it doesn't meet the conditions.
Moving to the next number, 11:
- The sum of its digits is 1 + 1 = 2, and the product is 1 × 1 = 1. Though 11 is divisible by 2, it is not divisible by 1, so it doesn't meet the conditions.
We continue this process for each number until we find the one we're looking for. I'll list a few more examples to demonstrate the steps:
12:
- Sum of digits = 1 + 2 = 3
- Product of digits = 1 × 2 = 2
- 12 is not divisible by 3 or 2.
13:
- Sum of digits = 1 + 3 = 4
- Product of digits = 1 × 3 = 3
- 13 is not divisible by 4 or 3.
14:
- Sum of digits = 1 + 4 = 5
- Product of digits = 1 × 4 = 4
- 14 is not divisible by 5 or 4.
Continuing this process, we eventually find the number that satisfies both conditions. I'll let you explore the rest of the numbers from 15 to 99 to see if you can find the answer.