Numbers that are divisible by both 3 and 8 must be divisible by their least common multiple, which is 24. Therefore, any number that is a multiple of 24 will be divisible by both 3 and 8. Examples of such numbers include 24, 48, 72, 96, and so on.
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
8 of the numbers less than 660 are divisible by 5 and 11 but not 3. All numbers divisible by 5 and 11 are multiples of their lcm; lcm(5, 11) = 55 All numbers divisibile by 5, 11 and 3 are multiples of their lcm; lcm(5, 11, 3) = 165 659 ÷ 55 = 11 54/55 → 11 numbers less than 660 are divisible by 5 and 11 659 ÷ 165 = 3 164/165 → 3 numbers less than 660 are divisible by 5, 11 and 3 → of the 11 numbers less than 660 divisible by 5 and 11, 3 are also divisible by 3 → 11 - 3 = 8 numbers less than 660 are divisible by 5 and 11 but not 3,.
No, its not divisible by 8. 639, is an odd numbers with the following divisors: 1, 3, 9, 71, 213, and itself.
Oh, dude, let me break it down for you. So, we start at 21 because it's the first number between 20 and 50 that's divisible by 3, and then we just keep adding 3 until we hit 48. So, that gives us 10 numbers in total. Easy peasy, right?
There are 41 numbers between 1 and 1,000 that are divisible by 3 and 8?
Numbers that are divisible by both 3 and 8 must be divisible by their least common multiple, which is 24. Therefore, any number that is a multiple of 24 will be divisible by both 3 and 8. Examples of such numbers include 24, 48, 72, 96, and so on.
36 is divisible by itself but not 8 36/36=1 36/8=4.5
Apply for the rules of 3 and 8. Numbers are divisible by 24 only if they are divisible by both 3 and 8.
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
11 is divisible by no whole numbers except 1 and 11.
5+2+1=8 and 8 is not divisible by 3.
Numbers divisible by 6 will have at least one 2 and one 3 as prime factors because 2 x 3 = 6. The same is true for 2, 4 and 8. Numbers divisible by 4 are also divisible by 2. Numbers divisible by 8 are also divisible by 4 and 2.
Prime numbers are only divisible by 1 and themselves. Since 3 is not a prime number, any prime number divisible by 3 does not exist. Therefore, the set of prime numbers divisible by 3 is an empty set.
Oh, dude, so like, numbers divisible by 6 are any numbers that can be divided by 6 without leaving a remainder. So, like, 6, 12, 18, 24, and so on are all divisible by 6 because 6 goes into them evenly. It's like the cool kid at the math party, just dividing everything effortlessly.
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
All numbers that are divisible by 8 and 12 would also be divisible by 4.