What are the divisibility rules for 96?
I don't know what you mean by visibility rules. The prime factors are
2*2*2*2*2*2*3,
It is evenly divisible by any combination of those factors, i.e.
2,4,8,16,32
3,6,12,24,48
To determine if a number is divisible by 96, you can apply the following divisibility rules:
1. Divisible by 2: A number is divisible by 2 if its last digit is divisible by 2. So, check if the units digit of the number is 0, 2, 4, 6, or 8.
2. Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. Add up all the digits of the number and check if the resulting sum is divisible by 3.
3. Divisible by 4: A number is divisible by 4 if the last two digits of the number together form a two-digit number that is divisible by 4. So, check if the last two digits of the number form a multiple of 4.
4. Divisible by 6: A number is divisible by 6 if it is divisible by both 2 and 3. So, apply the divisibility rules for 2 and 3 to determine if the number meets both conditions.
5. Divisible by 8: A number is divisible by 8 if the last three digits of the number together form a three-digit number divisible by 8. So, check if the last three digits of the number form a multiple of 8.
Now, let's apply these divisibility rules to the number 96:
1. Divisible by 2: Since the units digit is 6 (an even digit), 96 is divisible by 2.
2. Divisible by 3: The sum of the digits is 9 + 6 = 15. Since 15 is divisible by 3, 96 is divisible by 3.
3. Divisible by 4: The last two digits are 96. Since 96 is divisible by 4, 96 itself is divisible by 4.
4. Divisible by 6: Since 96 is divisible by both 2 and 3, it is divisible by 6 as well.
5. Divisible by 8: The last three digits are 096. Since 96 is divisible by 8, 96 is also divisible by 8.
Therefore, 96 is divisible by 2, 3, 4, 6, and 8.