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Q: What is the divisibility rules for 4and8and give one example for each?
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What is the prime factorization of 225 using the divisibility rules?

With the common divisibility rules, you can quickly see that it is divisible by 5, and by 9 (3 x 3). If you divide 225 by each of these numbers, you should be able to get the remaining factors quickly, as well.


How many days were in February in 1992?

Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)Since 1992 is divisible by 4, it is a leap year - meaning it has a February 29.(About divisibility by 4, special rules apply at the end of each century.)


What is the real divisibility rule for 9?

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)


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What are the Divisibility rules of 14 15 18?

14...Checking for divisibility by 14 would combine the rules for 2 and 7, since 2x7 = 14The rule for 2 is that the last digit of the number is even (0,2,4,6 or 8).There is no real rule for 7, but there is one way of testing divisibility by 7 that I have come up with, although rather complicated. First, write your number, then beneath it, under each digit from right to left, write out the numbers 1,3,2,6,4,5 in that order and keep repeating the sequence until each digit of your number has another digit beneath it. Use only the amount of digits needed. Then, multiply each digit of your number by the digit beneath it, and add all of these new numbers. If the number you get is a multiple of 7, the number you began with was also a multiple of 7.Example for 7:We are checking 3192847 for divisibility by 7...3 1 9 2 8 4 71 5 4 6 2 311x3 + 5x1 + 4x9 + 6x2 + 2x8 + 3x4 + 1x7= 3 + 5 + 36 + 12 + 16 + 12 + 791We can test 91 again to find divisibility...9 13 13x9 + 1x1 = 2828 is a multiple of 7 (it is 4 x 7), so 91 and 3192847 must also be multiples of 7. 3192847 is actually 456121 x 7.If both of these tests work, the number must be a multiple of 14.15...Divisibility by 15 can be tested by combining the rules for 3 and 5.The rule for 3 is that the sum of all the digits in the number is a multiple of 3.The rule for 5 is that the number ends in 5 or 0.If both of these tests work, the number must be a multiple of 15.18...Checking for divisibility by 18 combines the divisibility rules for 2 and 9.Once again, a number is divisible by 2 if its last digit is even.A number is divisible by 9 if the sum of its digits is a multiple of 9.If both of these tests work, the number must be divisible by 18.


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What is 8487 divisibility rule?

The number must be divisible by 9, 23 and 41, so all three of the following conditions must be met.Divisibility by 9 requires you to check the sum of the digits is divisible by 9.Divisibility by 23 requires you to add 7 times the last digit to the rest. The answer must be divisible by 23 directly.For divisibility by 41, subtract 4 times the last digit from the rest. The answer must be divisible by 23 directly.In each case, the additions or subtractions can be repeated so as to make the answer easier to test for divisibility.


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