If the number 715_423 is divisible by 3 (_ denotes the missing digit in the thousandths place), then the smallest whole number in the place of _ is _______.
2
The divisibility rule for the number 3 states that a whole number is divisible by 3 if the sum of its digits is divisible by 3. This rule is very useful for quickly checking if a number, no matter how large, can be evenly divided by 3 without performing long division.
We are given the number 715_423, where the underscore _ represents a missing digit. According to the question, this missing digit is in the thousands place. We need to find the smallest whole number that can replace this underscore so that the entire number 715_423 becomes divisible by 3.
First, let's represent the missing digit with a variable, say \(x\). The number can be written as \(715x423\).
Now, let's find the sum of all the digits in the number 715x423:
So, the sum of the digits of the number 715_423 is \(22 + x\).
For the number 715_423 to be divisible by 3, the sum of its digits, which is \(22 + x\), must also be divisible by 3.
Since \(x\) is a single missing digit, it must be a whole number from 0 to 9. We need to find the smallest whole number for \(x\) that makes \(22 + x\) divisible by 3. Let's test the possible values for \(x\) starting from 0:
| Value of \(x\) (Missing Digit) | Sum of Digits (\(22 + x\)) | Divisible by 3? |
|---|---|---|
| 0 | \(22 + 0 = 22\) | No (22 ÷ 3 is not a whole number) |
| 1 | \(22 + 1 = 23\) | No (23 ÷ 3 is not a whole number) |
| 2 | \(22 + 2 = 24\) | Yes (24 ÷ 3 = 8) |
From the table, we can see that when \(x = 2\), the sum of the digits is \(24\), which is perfectly divisible by 3. Since we started checking from the smallest possible whole number (0) for \(x\), the value 2 is the smallest whole number that satisfies the condition.
Therefore, the smallest whole number that can replace the underscore _ to make the number 715_423 divisible by 3 is 2. The complete number would then be 7,152,423.
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