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Q: What numbers are divisible by 2 3 5 and 9?

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Odd numbers are integers (whole numbers) that are not divisible by 2: 1, 3, 5, 7, ..., -1, -3, -5, -7...Odd numbers are integers (whole numbers) that are not divisible by 2: 1, 3, 5, 7, ..., -1, -3, -5, -7...Odd numbers are integers (whole numbers) that are not divisible by 2: 1, 3, 5, 7, ..., -1, -3, -5, -7...Odd numbers are integers (whole numbers) that are not divisible by 2: 1, 3, 5, 7, ..., -1, -3, -5, -7...

There are no such numbers.

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.

2: 1 and 2 3: 1 and 3 5: 1 and 5

No odd numbers are evenly divisible by 2. 3 ÷ 2 = 1 1/2 11 ÷ 2 = 5 1/2

no. in order for a number to be divisible by 3 its numbers have to have a sum of a number divisible by 3. EX. (302) 3+0+2 = 5. 5 is not divisible by 3. Therefore 302 is not divisible by 3.

It is divisible by all of those numbers.

No. 50 is divisible by these numbers: 1, 2, 5, 10, 25, 50.

yes it is divisible by all of those numbers

There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.

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

Yes. The prime factorization of 75 is 3 x 5 x 5. So 75 is divisible by the two prime numbers 3 and 5.

Eight is divisible, out of these numbers, by only 2.

90 is composite. 90 = 2 * 3 * 3 * 5 = 2 * 32 * 5

Infinitely many.

Multiples of 60.

6 (15,30,45,60,75,90)

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

Last digit (0) is even, so it is divisible by 2 4 + 3 + 2 + 0 = 9 which is divisible by 3, so it is divisible by 3 last digit is 0 or 5, so it is divisible by 5 4 + 3 + 2 + 0 = 9 which is divisible by 9, so it is divisible by 9 last digit is 0, so it is divisible by 10 → 4320 is divisible by all the numbers 2, 3, 5, 9, 10

because 10 is also divisible by 2 and 5.

None of the numbers you listed are evenly divisible by 1000.

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

No, but I suspect you want to know if 186624 is divisible by those numbers. Out of that list, 186624 is divisible by all but 5.

No. It is divisible by these numbers: 1, 2, 5, 10, 79, 158, 395, 790.

Prime numbers are only divisible by 1 and itself... so no prime number can be divisible by the numbers you listed.