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Hi Miska, To find the factors you first need to find the prime factors and to do that you need to find the prime numbers that divide 111. You only need to try the prime numbers up to the square root of 1111. The square root of 111 is a little more than 10 so you only need to try the prime numbers up to 10. There are two primes that are easy, 2 and 5. For a number to be divisible by 2 its final digit must be even. Thus the number 273 is not divisible by 2 since 3 is not even. For a number to be divisible by 5 its final digit must be 0 or 5. Hence 273 is not divisible by 5 either. There is also a simple rule for divisibility by 3. Take the number and add its digits. If the sum of the digits is divisible by 3 then the number is also divisible by 3. For example the sum of the digits of 273 is 2 + 7 + 3 = 12 and 12 is divisible by 3 so 273 is divisible by 3. Try it, divide 273 by 3. This rule for divisibility by 3 also holds for divisibility by 9. Take the number and add its digits. If the sum of the digits is divisible by 9 then the number is also divisible by 9. Now try 111. Penny | ||||||||||||
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