THE NUMBER 567 IS DIVISBLE BY WHICH OF THESE NUMBERS
2
5
9
10
Run them through a calculator and see what you find. Just looking at 567, I can immediately see 3 of the 4 numbers that cannot be divided evenly into it.
divisible by 9
To be divisible by 2, it must be an even number. To be divisible by 5, it needs to end in either a 5 or 0. To be divisible by 10, it needs to end with a 0.
What does that leave you?
I hope this helps a little more. Thanks for asking.
To determine whether the number 567 is divisible by each of the given numbers (2, 5, 9, 10), we can use the concept of divisibility rules. Here's what you need to know:
1. Divisible by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).
- The last digit of 567 is 7 (which is not even), so 567 is not divisible by 2.
2. Divisible by 5: A number is divisible by 5 if its last digit is 5 or 0.
- The last digit of 567 is 7 (which is not 5 or 0), so 567 is not divisible by 5.
3. Divisible by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
- The digits of 567 are 5, 6, and 7. Sum of these digits: 5+6+7=18. Since 18 is divisible by 9, we can conclude that 567 is divisible by 9.
4. Divisible by 10: A number is divisible by 10 if its last digit is 0.
- The last digit of 567 is 7 (which is not 0), so 567 is not divisible by 10.
Therefore, out of the given numbers, 567 is only divisible by 9.