What is a number greater than 300 that has an odd number of factors?

Answer 1

Only perfect squares have an odd number of factors. Generally factors pair up (e.g. factors of 12 are 1x12, 2x6, 3x4). But in the case of perfect squares, one number "pairs up" with itself, making for an odd number of factors.

In your case there are lots of perfect squares greater than 300. Assuming they want the first number with this characteristic, I'd pick 324. (324 = 18²).

The prime factors of 324 are:
2 x 2 x 3 x 3 x3

This means the factors of 324 are:
1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162 and 324

See how they pair up:
1 x 324 = 324
2 x 162 = 324
3 x 108 = 324
4 x 81 = 324
6 x 54 = 324
9 x 36 = 324
12 x 27 = 324
except for 18 which "pairs" with itself:
18 (18 x 18 = 324)

So 324 has an odd number of factors.

(Other possible answers are 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, etc.)

(Note: if you want to prove it simply, pick 361... the square root is prime, so it has only three factors of 1, 19 and 361.)

Answer 2

Squares, fourth powers, etc. all even powers.
I gave a detailed proof why, using the prime factorisation, in the second answer of http://answers.yahoo.com/question/index;_ylt=AlgXRdqQqxxb8yCsfaOm42Dsy6IX?qid=20061001165941AAsf6Po

( Do a prime decomposition of x:
x = p_1^alpha_1 . p_2^alpha_2 ... p_i^alpha_i
where its distinct prime factors are p_1, p_2, ... p_i and their multiplicities are alpha_i . See answer for details)

So pick from 18^2=324, 19^2=361, 20^2=400 ...
Personally I would pick 101^2 = 10201 or 111^2=12321 to prove the odd point :)
The factors of 10201 are 1, 101, 10201

The way to prove the number of factors and automatically generate all of them is to use the prime decomposition.
18=2.3^2 ; write the powers alpha_i as {1,2}
18^2=2^2.3^4 ; write the powers alpha_i as {2,4}

Then all the (2+1)*(4+1)=15 factors of 18^2 are generated by
(0,0)->1, (0,1) ->3, (0,2)->9, (0,3)->27, (0,4)->81
(1,0)->2, (1,1) ->6, (1,2)->18, (1,3)->54, (1,4)->162
(2,0)->4, (2,1) ->12, (2,2)->36, (2,3)->108, (2,4)->324

Computing factors of 18^2 reduced to just choosing unique tuples (0..2, 0..4) of which there are obviously (2+1)*(4+1).
Similarly for any other even power >300.

Answer 3

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