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Sagot :
Sure! Let's break down the problem step by step and find the solutions given the specific requirements for each number and condition mentioned:
### Given Numbers:
1. The number 5
2. The number 40
3. The number 21
4. The number 21
5. The number 3231 with a square to fill in at last place
Now, let's check the divisibility rules for each case:
### 1. Divisibility by 2 and 5:
- 2: A number is divisible by 2 if its last digit is even (i.e., 0, 2, 4, 6, 8).
- 5: A number is divisible by 5 if its last digit is either 0 or 5.
### Detailed Analysis:
Let's focus on the 5th case that describes the number 3231 with a digit to fill in at the last place. The task is to identify which digit makes 3231[tex]$\square$[/tex]:
1. Divisible by 10:
- For a number to be divisible by 10, its last digit must be 0.
- Therefore, 3231 appended with 0 (32310) is fully divisible by 10 and hence also divisible by both 5 and 2.
2. Divisible by 5:
- For a number to be divisible by 5, the last digit must be 0 or 5.
- Therefore, 32315 (adding digit 5) also makes the number divisible by 5.
By checking all digits (0 to 9) for 3231[tex]$\square$[/tex], we get:
Filling the digit 0 gives 32310 which is divisible by both 5 and 2.
Filling the digit 1 gives 32311, not divisible by 5 or 2.
Filling the digit 2 gives 32312, divisible by 2 but not by 5.
Filling the digit 3 gives 32313, not divisible by 2 or 5.
Filling the digit 4 gives 32314, divisible by 2 but not by 5.
Filling the digit 5 gives 32315, divisible by 5 but not by 2.
Filling the digit 6 gives 32316, divisible by 2 but not by 5.
Filling the digit 7 gives 32317, not divisible by 2 or 5.
Filling the digit 8 gives 32318, divisible by 2 but not by 5.
Filling the digit 9 gives 32319, not divisible by 2 or 5.
### Conclusion:
Divisible by 10 (implies divisible by both 2 and 5):
- The number 3231[tex]$\square$[/tex] with 0 makes it 32310.
Divisible by 5:
- The number 3231[tex]$\square$[/tex] with either 0 or 5 (i.e., 32310 or 32315).
### Additional Notes:
1. For any specific missing digit in 3231[tex]$\square$[/tex]:
To be divisible by 2, end with 0, 2, 4, 6, or 8.
To be divisible by 5, end with 0 or 5.
* To be divisible by both 5 and 2 (which is essentially divisible by 10), end with 0.
The correct digit making 3231[tex]$\square$[/tex] divisible by 5 and 2 specifically is:
0 (32310).
That means:
- The number 5 is irrelevant here.
- The number 40 is divisible by 2 and 10.
- The number 21 is only divisible by 3 and 7, but that's irrelevant here.
- The number 3231 should end with 0 to be divisible by both 2 and 5, i.e., 32310.
So, the filled number pattern:
3231[tex]$\square$[/tex] becomes 32310.
### Given Numbers:
1. The number 5
2. The number 40
3. The number 21
4. The number 21
5. The number 3231 with a square to fill in at last place
Now, let's check the divisibility rules for each case:
### 1. Divisibility by 2 and 5:
- 2: A number is divisible by 2 if its last digit is even (i.e., 0, 2, 4, 6, 8).
- 5: A number is divisible by 5 if its last digit is either 0 or 5.
### Detailed Analysis:
Let's focus on the 5th case that describes the number 3231 with a digit to fill in at the last place. The task is to identify which digit makes 3231[tex]$\square$[/tex]:
1. Divisible by 10:
- For a number to be divisible by 10, its last digit must be 0.
- Therefore, 3231 appended with 0 (32310) is fully divisible by 10 and hence also divisible by both 5 and 2.
2. Divisible by 5:
- For a number to be divisible by 5, the last digit must be 0 or 5.
- Therefore, 32315 (adding digit 5) also makes the number divisible by 5.
By checking all digits (0 to 9) for 3231[tex]$\square$[/tex], we get:
Filling the digit 0 gives 32310 which is divisible by both 5 and 2.
Filling the digit 1 gives 32311, not divisible by 5 or 2.
Filling the digit 2 gives 32312, divisible by 2 but not by 5.
Filling the digit 3 gives 32313, not divisible by 2 or 5.
Filling the digit 4 gives 32314, divisible by 2 but not by 5.
Filling the digit 5 gives 32315, divisible by 5 but not by 2.
Filling the digit 6 gives 32316, divisible by 2 but not by 5.
Filling the digit 7 gives 32317, not divisible by 2 or 5.
Filling the digit 8 gives 32318, divisible by 2 but not by 5.
Filling the digit 9 gives 32319, not divisible by 2 or 5.
### Conclusion:
Divisible by 10 (implies divisible by both 2 and 5):
- The number 3231[tex]$\square$[/tex] with 0 makes it 32310.
Divisible by 5:
- The number 3231[tex]$\square$[/tex] with either 0 or 5 (i.e., 32310 or 32315).
### Additional Notes:
1. For any specific missing digit in 3231[tex]$\square$[/tex]:
To be divisible by 2, end with 0, 2, 4, 6, or 8.
To be divisible by 5, end with 0 or 5.
* To be divisible by both 5 and 2 (which is essentially divisible by 10), end with 0.
The correct digit making 3231[tex]$\square$[/tex] divisible by 5 and 2 specifically is:
0 (32310).
That means:
- The number 5 is irrelevant here.
- The number 40 is divisible by 2 and 10.
- The number 21 is only divisible by 3 and 7, but that's irrelevant here.
- The number 3231 should end with 0 to be divisible by both 2 and 5, i.e., 32310.
So, the filled number pattern:
3231[tex]$\square$[/tex] becomes 32310.
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