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Q: How does the divisibility rule for 4 and 8 work?

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There isn't an especially easy divisibility rule for 8, but because 1,000 divides by 8, you can say any digits greater than the hundreds are irrelevant for divisibility. So to work out if a number is divisible by 8, you only need to look at the last 3 numbers. Divide these numbers by 4, and if the result is even, that means the whole number is divisible by 8.

What is the divisblity rule by 8

That the last digit is 0. 2 4, 6 or 8.

The divisibility rule for 22 is that the number is divisible by 2 and by 11. Divisibility by 2 requires that the number ends in 0, 2, 4, 6 or 8. Divisibility by 11 requires that the difference between the sum of the the digits in odd positions and the sum of all the digits in even positions is 0 or divisible by 11.

The number formed by the last three digits have to be divisible by 8.

A divisibility rule is a statement of procedures to determine divisibility. For example:"A number is divisible by 2 if it ends with 0, 2, 4, 6 or 8".It does not make any sense to divide such a sentence by 2 or 895 or any other number!

If the number is even it is divisible by 2. The number is even if its units digit is even, that is, if it is 0, 2, 4, 6 or 8.

4- last two digits are a multiple of 4 7-take the last digit and subtract it from the rest of the number and if multiple of 7 (including 0) 8- last three digits are multiple of 8

If it ends in 2, 4, 6, 8, or 0 (zero), then it's divisible by 2.If it's an even number, it's divisible by 2.

A number is divisible by 8 if and only if it is an integer and the last three digits make a number that is divisible by 8. A general divisibility rule for 2^n where n is an integer is to take the last n digits and if they make a number that is divisible by n then it is divisible by n.

If the last digit (6) is divisible by 2 - that is, it is 0, 2, 4, 6 or 8 then the number is divisible by 2. Note that this rule cannot be used to check for divisibility by other numbers.

The divisibility rule for 42 isdivisibility by 2, ANDdivisibility by 3, ANDdivisibility by 7.Divisibility by 2 requires the number to be even. That means it must end in 0, 2, 4, 6 or 8.Divisibility by 3 requires that the digital root of the number is divisible by 3. That is, the sum of the digits is divisible by 3. If the first sum is large, you can calculate the digital root of the digital root (and again, if necessary) and check that for divisibility by 3.The divisibility rule for 7 is more difficult.Take the last digit (units).From the number formed by the remaining digit, subtract twice the last digit.If the answer is divisible by 7 (including zero or negative numbers), then the original number is divisible by 7.For large numbers, repeat the process to bring the number down to a manageable size.

The divisibility rule of 3 is that you add the digits together and then if the number you come up with is divisible by 3, then the number itself is divisible by 3. In this instance, 5 + 8 + 5 = 18/3 = 6, therefore, 585 is divisible by 3.

It would be 9,992 since 1,249 * 8 = 9992. 1250 * 8 would equal 10,000 and this is not a 4 digit number, but a 5 digit number. The divisibility rule for 8 states that the last 3 digits of a number must be divisible by 8 and this is true as 992/8 = 124. Hopefully that helps.

No.

if the last 3 digits are divisible by 8,so is the number. Ex.: 9 is not because 009 is not. 6008 is because 008 is.

There are two parts to the rule: Part 1: If the last but one digit is even, the last digit must be 0, 4 or 8 If the last but one digit is odd, then the last digit must be 2 or 6. Part 2: Find the sum of the first, third, fifth, etc digits. Find the sum of the second, fourth, sixth, etc digits. The difference between these two sums must be a multiple of 11 ie 0, 11, 22, 33 etc. Part 1 ensures divisibility by 4 Part 2 ensures divisibility by 11 Together, they ensure divisibility by 44.

If the last two digits are divisible by 4. This is equivalent to: Last digit = 0, 4, 8 and the digit before (in tens place) is even or Last digit = 2 or 6 and the digit before (in tens place) is odd.

Those for 1, 2, 4, 5 and 8.

You know from the rules of divisibility that 10000 is divisible by 2, 4, 5, 8 and 10. Divide it by one of those. The integer you come up with will create a factor pair of 10000.

Any multiple of two must end in 0, 2, 4, 6 or 8.

In a calculater or by hand do 295/8 OR by using the divisibility rule of 8 if the last 3 digits are divisible by 8, so is the entire number. So you would still do 295/8 but now you know why.

In most cases, it is easier to do the division than to use some special divisibility rule. Divisibility rules only save you time in some special cases, such as:2, 4, 8, 5, 10, 25 (look at the last digit or digits)3, 9 (add the digits, and see if it is divisible by 3 or by 9, respectively)7, 11, and 13 (separate the last three digits and subtract from the remaining number)

When you add each digit, it's always equals to 9 or a multiple of 9. Example 18 (1+8=9) , 27 (2+7=9)

Do the division, if there is no remainder, it is divisible. Seriously, many of the "divisibility rules" that have been discovered become more complicated than doing the actual division. For practical purposes, just learn the divisibility rules for a few simple cases (divisibility rules by 2, 4, 8, 5, 10, 3, 9, 7, 11, and 13), and for all other cases, just do the division.