# why do square numbers have odd factors Let's consider a few cases: \$n\$ is even/odd and \$n\$ is an odd square or not. Suppose that \$n\$ is odd and is not a square. Then, each of the factors of \$n\$ can be paired with one another. More precisely, \$a\$ is paired with \$b\$ if \$n=ab\$. Since \$n\$ is odd, both \$a\$ and \$b\$ are odd, but their sum is even. Therefore, the sum of the factors of \$n\$ is even. Suppose that \$n\$ is odd and is a square. Then, we can pair each of the factors of \$n\$ as above, except for the square root. This leaves several pairs of odd integers, and one integer left over. Therefore, the sum of the factors of \$n\$ is odd. Suppose that \$n\$ is even and can be written as \$n=2^km\$ where \$m\$ is odd. The odd factors of \$n\$ are precisely the odd factors of \$m\$. Therefore, we can use the cases above to determine if the sum is odd or even. Combining these, the odd squares are: 1, 9, 25, 49, 81. Now, we consider powers of two times these to get: 2, 4, 8, 16, 32, 64, 18, 36, 72, 50, 98. These are the 16 possibilities less than 100
We have two responses for you. Hello Jacqueline. The short answer to your question is that they don't. Only the numbers that are perfect squares have an odd number of factors. For example, the factors of 16 are 1, 2, 4, 8, 16. Pairs of factors multiplied together give 16: 1x16, 2x8 and 4x4. Since they are paired, there is an even number, but we don't list the same number twice, so 16 has 5 factors rather than 6. Compare that to 48, whose factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Pairs are 1x48, 2x24, 3x16, 4x12, 6x8. That's an even number of factors (no factor appears twice). Stephen La Rocque. > Actually 48 has an even number of factors as do 60 and 90. The correct question is, maybe, why is it only the square numbers, 1,4,8,15,25,. have an odd number of factors? The basic idea is as follows: consider 60 as a product of primes, i. e. , 60 = 2^2x3x5. If you are to 'build' a factor of 60 you may include 0, 1 or 2 2's, 0 or 1 3's, and 0 or 1 5's, all told 3 choices followed by 2 choices followed by 2 choices which leads to 12 factors, an even number of them. The result is always even whenever some of the prime factors of the number that you are looking at appear to an odd power (as 3 and 5 did above) because this gives you an even number of choices for that prime when you're building your factors. The squares have only prime factors to even powers and thus the number of choices for a particular prime factor, when building factors, is odd in each case. To really see this, try writing the number in your list as products of primes first and then try to build the factors as suggested above.

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