Maud from Hymers College in England and Luke from Long Field Academy, both in England, used Alison's method to count the factors of 360. Maud wrote:
To find the factors you need to know what factors are, factors can't be above the number you are trying to find the factors for, so you would divide 360 by a number (I would start at 1 then keep going up) and that would be 360 so 360 and 1 are both factors and you would do this for the rest until it repeats itself.
Holly and Amy from Long Field Academy and Sanika from India used Charlie's method to count how many factors 360 has. This is Holly's work:
$360 = 2^3\times3^2\times5^1$
Number of factors $=4\times3\times2 = 24$
Alex and Amy from Long Field Academy and Sanika also used Charlie's method to show that Alison and Charlie's other numbers have 24 factors. This is Alex's work:
25725= 5^{2}$\times$3^{1}$\times$7^{3}
Click to see Alex's table
5^{0} 
3^{0} 
7^{0} 
1 
7^{1} 
7 

7^{2} 
49 

7^{3} 
343 

3^{1} 
7^{0} 
3 

7^{1} 
21 

7^{2} 
147 

7^{3} 
1029 

5^{1} 
3^{0} 
7^{0} 
5 
7^{1} 
35 

7^{2} 
245 

7^{3} 
1715 

3^{1} 
7^{0} 
15 

7^{1} 
105 

7^{2} 
735 

7^{3} 
5145 

5^{2} 
3^{0} 
7^{0} 
25 
7^{1} 
175 

7^{2} 
1225 

7^{3} 
8575 

3^{1} 
7^{0} 
75 

7^{1} 
525 

7^{2} 
3675 

7^{3} 
25725 
So there will be 3 rows in the first column which match the 5. Also in the second column there will be 2 rows to match the 3. And there will be 4 that match the 7. The number of factors = 3 $\times$ 2 $\times$ 4 = 24.
217503 = 11^{1}$\times$13^{3}$\times$3^{2}
So there will be 2 rows in the first column which match the 11. Also in the second column there will be 4 rows to match the 13. And there will be 3 that match the 3. The number of factors = 2 $\times$ 4 $\times$ 3 = 24.
312500 = 5^{7} $\times$ 2^{2}
So there will be 8 rows in the first column which match the 7. Also in the second column there will be 3 rows to match the 2. The number of factors = 8 $\times$ 3 = 24.
690625 = 17^{1}$\times$13^{1}$\times$ 5^{5}
So there will be 2 rows in the first column which match the 17. Also in the second column there will be 2 rows to match the 13. And there will be 6 that match the 5. The number of factors = 6 $\times$ 2 $\times$ 2 = 24.
94143178827 = 3^{23}
So there will be 24 rows in the first column which match the 3. The number of factors = 24 $\times$ 1 = 24.
Sanika and Amy found the smallest numbers with 14 and 15 factors. This is Amy's work:
To find the smallest number with 14 factors, you would list all of the factors of 14:
2 $\times$ 7
1 $\times$ 14
Then, you would pick the pair of numbers that have the least difference between them. In this case, it is 2 and 7. Then take 1 away from each of the numbers:
2$$1 = 1
7$$1 = 6
These numbers will be the powers. Then, you would pick the two smallest prime numbers, however, they must be different else the number of factors will be incorrect.
The two prime numbers that you would need to use are 2 and 3. Then, you would give the smallest prime number (2), the biggest power (6).
You need to do this in order to ensure that you get the smallest number possible.
So, the number would be:
2^{6} $\times$ 3^{1} = 192
This method would be the same to work out the smallest number of factors for any number, all that you need to change would be to swap the number 14 with the number of factors that you want your number to have. For example,
to find the smallest number with 15 factors:
Factors of 15:
3 $\times$ 5
1 $\times$ 15
Choose the pair of factors with the least difference:
3 and 5
Take away 1:
3$$1 = 2
5$$1 = 4
Smallest, different, prime numbers:
2 and 3
Powers:
2^{4} $\times$ 3^{2}
2^{4} $\times$ 3^{2} = 144
This is Sanika's work for a number with 18 factors:
To find numbers with exactly 18 factors we must make sure the product of all the exponents (+1) of the prime factors give us 18. The factors of 18 include (1,18) (2,9) (3,6). We can further prime factorise the last two pairs to give us (2, 3, 3). Smaller the exponent the lesser the value is going to be. So we will take (2, 3, 3) as the exponents. We will take the first three primes as bases i.e. 2 3 & 5. This will give us a final answer of 2^{2}$\times$3^{2}$\times$5. This leaves us with 180.
Sanika also explained which numbers have an odd number of factors:
Numbers which are perfect squares will have an odd number of factors as one number will be repeated whilst writing down its factor pairs.
E.g. Let’s consider 16= 2^{4} $\rightarrow$ (1, 16) (2, 8) (4, 4)
So 16 has a total of 5 factors as 4 is repeated twice. This holds true for all other perfect squares.
Sanika found the smallest number with exactly 100 factors:
Prime factorising 100 gives us 2$\times$2$\times$5$\times$5, now we can assign the smallest prime bases to the largest powers, which will give us 2^{4}$\times3^{4}$\times$5$\times$7 = 45360.
Sanika and Amy both said that the number under 1000 with the most factors is 840 (32 factors). This is Amy's work:
Finally, to find the number less than 1000 that has the most factors, you have to try and multiply together the widest range of prime numbers, starting with the lowest:
2 $\times$ 3 $\times$ 5 $\times$ 7 $\times$ 11 $\gt$ 1000 but,
2 $\times$ 3 $\times$ 5 $\times$ 7 $\lt$ 1000 so these are the numbers that we will use.
2 $\times$ 3 $\times$ 5 $\times$ 7 = 210, so, now we can add powers to the number 2 until the number is as close to 1000 as possible.
Therefore, the answer is 2^{3} $\times$ 3^{1} $\times$ 5^{1} $\times$ 7^{1} = 840: this is the closest number to 1000.
Using the information above, the number of factors that it has must be 32.