Friday

November 28, 2014

November 28, 2014

Posted by **chris** on Tuesday, December 19, 2006 at 3:27pm.

Odd numbers can only be divided evenly by another odd number.

Even numbers can be evenly divided by either odd or even numbers.

2---Numbers that end in 0, 2, 4, 6, or 8 are evenly divisible by 2.

3---If the sum of a number's digits is evenly divisible by 3, the number is divisible by 3.

4---If the last two digits are both zero or they form a two digit number evenly divisible by 4, then

the whole number is evenly divisible by 4.

5--Any number ending in 5 or 0 is evenly divisible by 5.

6--If the number is evenly divisible by 2 and 3, it is divisible by 6.

If the number is even and the sum of the digits is evenly divisible by 3, the whole number is

divisible by 3.

7---Double the last digit and subtract from the number without the last digit. If the result is evenly

divisible by 7, so is the original number.

Multiply the left-hand digit by 3 and add the next digit. With the result, repeat as often as

necessary with the subsequent digits. If the final answer is divisible by 7, so is the original

number.

Divide the number into groups of 3 digits or thousands. Subtract the first group from the

second group. Subtract the third group from the result of the first subtraction. Continue

subtracting until the last group is subtracted. If the resulting number is divisible by 7, so is the

original number. This method works for 11 and 13 as well.

8---If the last 3 digits are zero or if they form a number that is evenly divisible by 8, then the whole

number is evenly divisible by 8.

9---If the sum of the digits is evenly divisible by 9, the number is evenly divisible by 9.

10--Any number that ends in 0 is evenly divisible by 10.

11--If all the digits in the number are the same and it has an even number of digits, the number is evenly

divisible by 11. An alternate method is to 1) add up the odd position digits, 2) add up the even position digits, and 3) calculate the difference between the two sums. If the difference is divisible by 11, the original number is divisible by 11.

Divide the number into groups of 3 digits or thousands. Subtract the first group from the

second group. Subtract the third group from the result of the first subtraction. Continue

subtracting until the last group is subtracted. If the resulting number is divisible by 11, so is the

original number.

12--If a number can be evenly divided by 3 and 4, then it can be evenly divided by 12.

13--Divide the number into groups of 3 digits or thousands. Subtract the first group from the

second group. Subtract the third group from the result of the first subtraction. Continue

subtracting until the last group is subtracted. If the resulting number is divisible by 13, so is the

original number.

15--If a number can be evenly divided by 3 and 5, then it can be evenly divided by 15.

17--Remove the last digit and multiply it by 5. Subtract the result from the rest of the number. Continue this process until you have a number smaller than 40. If this final result is divisible by 17, then the original number is divisible by 17.

18--If the number is evenly divisible by either 2 and 9 or 3 and 6, the number is divisible by 18.

**Answer this Question**

**Related Questions**

math problem - My ID number is quite remarkable.Its a 9 digit number with the ...

Statistics - a roulette wheel has 40 slots evenly divided between red (even) and...

math - Jennifer considered the numbers 4, 24, 44, and 64 and noticed that they ...

math 156 - Jennifer considered the numbers 4, 24, 44, and 64 and noticed that ...

Reposted math question - "what do you call numbers that cannot be arranged into ...

Math - how do you know if one is a factor of another without dividing. a) 4320 ...

Math - 1. What is the 117th odd natural number? 2. What is the greatest three-...

Geometry - Working on conjectures. The question is Conjecure: The product of any...

math - Consider the following sums of numbers and how they are formed: 1 odd ...

Mathematics - Cardinal Number - Hi, This is a really easy question. I think I am...