Q: What number is divisible by 3 or 4 that gives you a remainder 20?

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Because the remainder when 20 is divided by any of them is 0.

No. To check for divisibility by 9 add the digits of the number and if the sum is divisible by 9 then the original number is also divisible by 9, otherwise it is not. For 24815: 2 + 4 + 8 + 1 + 5 = 20 which is not divisible by 9, so 24815 is not divisible by 9. Note that the test can be applied to the sum: repeatedly summing the digits of the sum of the digits until a single digit remains means the original number is divisible by 9 only if this single digit is 9, otherwise it is not (instead it gives you the remainder when the original number is divided by 9). This single digit is known as the digital root of the number.

No, 81 is not divisible by 4. In order for a number to be divisible by 4, the last two digits must be divisible by 4. In this case, the last two digits of 81 are 81 itself, which is not divisible by 4. Therefore, 81 is not divisible by 4.

since 648 = 2334 648 is divisible by any number that is 2x3y where x is between 0 and 3 and y is between 0 and 4. that gives you 20 possible answers.

640.

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A number is divisible by another number is, after dividing by that number, the answer is an integer with no remainder left over. For this reason, no number can be divisible by a number greater than itself. Thus, neither 1, nor 2, 3, 5, 8 or 20 are divisible by 15,345.

The meaning of "can be divided by" has lost a bit of its precision, since (with the exception of zero) anything can be divided by anything. Some people use it interchangeably with "divisible," which in mathematics means that there will be no remainder. Example: 20 can be divided by 8. The answer is 2.5 20 is not divisible by eight since the answer isn't a whole number. 20 can be divided by 10. The answer is 2. 20 is divisible by 10 since there isn't a remainder.

Because the remainder when 20 is divided by any of them is 0.

No. To check for divisibility by 9 add the digits of the number and if the sum is divisible by 9 then the original number is also divisible by 9, otherwise it is not. For 24815: 2 + 4 + 8 + 1 + 5 = 20 which is not divisible by 9, so 24815 is not divisible by 9. Note that the test can be applied to the sum: repeatedly summing the digits of the sum of the digits until a single digit remains means the original number is divisible by 9 only if this single digit is 9, otherwise it is not (instead it gives you the remainder when the original number is divided by 9). This single digit is known as the digital root of the number.

The smallest number that is divisible by 20 and 15 is 60.

They are: 5, 7, 11, 13, 17 and 19 which are all divisible by 4 but they all will have a remainder.

20

35 - 18 = 17 17/n = 20 n = 17/20 = 0.85

Yes, if the number is divisible by 20 then it has to be even. Typically if a number is divisible by any number ending in zero (10, 20, 30, etc), the number itself ends in zero as well, which makes it even.

If a number is divisible by both 4 and 5, it's divisible by 20.

if 4 multiplies into that number e.g. 4x4=16 so 16 is divisible by 4, 4x5=20 so 20 is divisible by 4

since 648 = 2334 648 is divisible by any number that is 2x3y where x is between 0 and 3 and y is between 0 and 4. that gives you 20 possible answers.