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Any multiples of 5 are exactly divisible by 5, i.e. any number 5 or larger that ends in 5 or 0 (zero). For example 2789645, 65, 8594620 and 950 are all divisible exactly by 5 as they are all multiples of 5, ending in either 5 or 0 (zero).

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βˆ™ 11y ago
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βˆ™ 11y ago

If a number is divisible by 4 and 5, it has to also be divisible by 20.Since 1 in 20 numbers are divisible by 20, the odds are 1/20, or 0.05.

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βˆ™ 8y ago

Any element of the set of numbers of the form 60*k, where k is an integer, is evenly divisible.

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βˆ™ 11y ago

20 and all its multiples.

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βˆ™ 8y ago

Any multiple of 20 from 1000 to 9980

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βˆ™ 8y ago

Any multiple of 60.

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βˆ™ 8y ago

It is divisible by 1, 2 and 4.

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βˆ™ 10y ago

30

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βˆ™ 8y ago

1, 2 and 4

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Q: What numbers are divisible by 2 3 4 and 5?
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What are the divisibility rules for 1 2 3 4 5 6 9 and 10?

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.


Why is that when a number is divisible by 2 and divisible by3 it is also by six how are they related why does this pattern not work for 2 4and 8?

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.


Is 244 divisible by 3?

No since only numbers in which its sum can be divisible by 3 is divisible by 3. 2 + 4 + 4 is 10, and 10 is not divisible by 3.


Is a number always divisible by 4 if the sum of the digits is 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


If a number is both divisible by 4 and 9 by what other numbers is it divisible?

1, 2 and 3


If a number is divisible by 12 what other numbers is it divisible by?

1, 2, 3, 4, and 6


How many 2 digit numbers are divisible by 3 and 4?

Eight of them.


What are the prime numbers between 2 and 10?

3, 5, 7. 4, 6, and 8 are divisible by 2. 9 is divisible by 3.


Is 524 divisibility by 3?

no.. heres a trick: if you add together the numbers (5+2+4) and that answer equals a number that is divisible by 3 then the number is divisible by three 5+2+4=11 11 is not divisible by 3 therefore, 524 is not divisible by 3


How is the two and four times tables connected?

2: 2,4,6,8,10,12,14,16,18,20...4: 4,8,12,16,20,24,28,32,36,40...Half of all numbers divisible by 2 are divisible by 4.All numbers divisible by 4 are divisible by 2.


Is 76 divisible by 2 3 4 5 6 9 and 10?

76 is divisible by some of those numbers, but not all of them. 76 is an even number that ends in 6. It is divisible by 2. The sum of the digits of 76 (13) is not evenly divisible by 3. 76 is not divisible by 3. 4 times 19 is 76. 76 is divisible by 4. Only numbers that end in 0 and 5 are divisible by 5. 76 is not divisible by 5. 6 and 9 do not divide evenly into 76, so 76 is not divisible by 6 and 9. Only numbers that end in 0 are divisible by 10. 76 is not divisible by 10. Of the numbers listed, 76 is only divisible by 2 and 4.


Are all numbers divisible by 3 and 4 divisible by 6?

Yes. If a number is evenly divisible by both 3 and 4, it will also be divisible by 6.This is because the prime factor of 3 is [3] -- in other words, 3 is a prime number -- and the prime factors of 4 are [2, 2].Thus, any number that is divisible by 3 and 4 will have as part of its prime factors the set [2, 2, 3]. (The smallest such number is 12, which is 2 x 2 x 3.) Since the prime factors of 6 are [2, 3] -- and [2, 3] is a subset of [2, 2, 3] -- then any such number must also be divisible by 6.Another way to look at it is to note that all numbers which are divisible by both 3 and 4 are also divisible by 12 (which is 3 x 4). Since 12 is divisible by 6, then all multiples of 12 will also be divisible by 6.