How does one go about finding all the factors of a natural number N? For small numbers, we simply factor N into its prime factors, and then take all possible combinations of these prime factors to determine its factors. Let’s consider natural numbers which are powers of a single prime.

For N = 7³, we get factors of 7, 7², and 7³ (3 factors), and, of course, a factor of 1 [7³ = 1(7³)]

For N = 13⁵, we get factors of 13, 13², 13³, 13⁴, and 13⁵ (5 factors), and, of course, a factor of 1

For N = pⁿ¹, we get factors of p, p², p³ . . . pⁿ¹ (n₁ factors), and, of course, a factor of 1

The number of factors T of a natural number N = pⁿ¹, where p is prime, is given by T = n₁ + 1.

Let’s now consider natural numbers which are the products of powers of two primes.

To find all of the factors of N = p⁵q², we select first all of the possible powers of each of the primes p and q taken one at a time (p, p², p³, p⁴, p⁵ and q, q²), giving us a total of 5 + 2 = 7 factors. Then, we select all of the possible products of the powers of the primes p and q taken two at a time, namely

pq and pq², p²q and p²q², p³q and p³q², p⁴q and p⁴q², p⁵q and p⁵q²,

giving us 5 groups of 2 factors each for a total of 5(2) = 10 more factors. Counting the factor 1, this gives us a grand total T of 5 + 2 + 5(2) + 1 = 18 factors.

To find all of the factors of N = pⁿ¹qⁿ² , we select first all of the possible powers of each of the primes p and q taken one at a time (p, p² . . . pⁿ¹ and q, q² . . . qⁿ²), giving us n₁ + n₂ factors. Then, we select all of the possible products of the powers of the primes p and q taken two at a time

pq, pq² . . . pqⁿ² and p²q, p²q² . . . p²qⁿ² , and . . . and pⁿ¹q, pⁿ¹q² . . . pⁿ¹qⁿ²,

giving us n₁ groups of n₂ factors each for a total of n₁n₂ more factors. Counting the factor 1, this gives us a grand total T of n₁ + n₂ + n₁n₂ + 1 = (n₁ + 1)(n₂ + 1) factors. Thus,

The number of factors T of a natural number N = pⁿ¹qⁿ², where p and q are primes,

Let’s now consider natural numbers which are the products of powers of three primes.

To find all of the factors of N = pⁿ¹qⁿ²sⁿ³ , we select first all of the possible powers of each of the primes

p, q, and s taken one at a time ( p, p² . . . pⁿ¹ and q, q² . . . qⁿ² and s, s² . . . sⁿ³), giving us

Next, we select all of the possible products of the powers of the primes p, q, and s taken two at a time.

a) Starting first with the products of the powers of p and q, we have

pq, pq² . . . pqⁿ² and p²q, p²q² . . . p²qⁿ² , and . . . and pⁿ¹q, pⁿ¹q² . . . pⁿ¹qⁿ²,

giving us n₁ groups of n₂ factors each for a total of n₁n₂ factors.

ps, ps² . . . psⁿ³ and p²s, p²s² . . . p²sⁿ³ , and . . . and pⁿ¹s, pⁿ¹s² . . . pⁿ¹sⁿ³,

giving us n₁ groups of n₃ factors each for a total of n₁n₃ factors.

qs, qs² . . . qsⁿ³ and q²s, q²s² . . . q²sⁿ³ , and . . . and qⁿ²s, qⁿ²s² . . . qⁿ²sⁿ³,

giving us n₂ groups of n₃ factors each for a total of n₂n₃ factors.

The total from a), b), and c) is n₁n₂ + n₁n₃ + n₂n₃ more factors. (2)

Finally, we select all of the possible products of the powers of the primes p, q, and s taken three at a

time. Since we have n₁ choices for the powers of p, n₂ choices for the powers of q, and n₃ choices for the

powers of s, we then, by the fundamental principle of counting, will have n₁n₂n₃ choices for the powers of

p, q, and s taken three at a time, giving us a total of n₁n₂n₃ factors. (3)

Combining the totals from (1), (2), and (3), and counting the factor 1, we get a grand total T of

(n₁ + n₂ + n₃) + (n₁n₂ + n₁n₃ + n₂n₃) + (n₁n₂n₃) + 1 = (n₁ + 1)(n₂ + 1)(n₃ + 1) factors. Thus,

The number of factors T of a natural number N = pⁿ¹qⁿ²sⁿ³, where p, q, and s are primes,

The number of factors of a natural number N = p₁ⁿ¹p₂ⁿ²p₃ⁿ³ . . . p_k^nk, where p₁, p₂, . . . , p_k

are primes, is given by T = (n₁ + 1)(n₂ + 1)(n₃ + 1) . . . (n_k + 1).

Example. How many factors does 720 have? Since 720 = 10(72) = 2(5)(8)(9) = 2(5)(2³)(3²) =

(2⁴)(3²)(5¹), its number of factors T is (4+1)(2+1)(1+1) = 5(3)(2) = 30. They are

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 90, 120,

From 2⁴(3²) 2(3), 2(3²), 2²(3), 2²(3²), 2³(3), 2³(3²), 2⁴(3), 2⁴(3²)

From 2⁴(3²)(5¹) 2(3)(5¹) 2²(3)(5¹) 2³(3)(5¹) 2⁴(3)(5¹)

Example. Find the smallest natural number N which has exactly 24 factors.

Now, N = p₁ⁿ¹p₂ⁿ²p₃ⁿ³ . . . p_k^nk, and to find N, we must find values for n₁, n₂, etc. so that N

will have exactly 24 factors. So T = 24, and from above, T = (n₁+1)(n₂+1) . . . (n_k+1). If we factor

n₁ + 1 = 8, and n₂ + 1 = 3, giving us n₁ = 7 and n₂ = 2.

Then, for any primes p₁ and p₂, the natural number N = p₁⁷p₂² will have exactly 24 factors.

To get the smallest N, we pick p₁ = 2 and p₂ = 3, the first two smallest primes, and we get

N = 2⁷3² = 128(9) = 1,152. But, is this really the smallest N? Let's factor T = 24 as 24(1),

a) For T = (n₁+1)(n₂+1) = 24(1), we get n₁ + 1 = 24, and n₂ + 1 = 1. So, n₁ = 23 and n₂ = 0.

Thus, N = p₁²³p₂⁰ = 2²³3⁰= 2²³ = 8,388,608, a very large number!

b) For T = (n₁+1)(n₂+1) = 12(2), we get n₁ + 1 = 12, and n₂ + 1 = 2. So, n₁ = 11 and n₂ = 1.

Thus, N = p₁¹¹p₂¹ = 2¹¹3¹ = 6,144, again a large number!

c) For T = (n₁+1)(n₂+1) = 6(4), we get n₁ + 1 = 6, and n₂ + 1 = 4. So, n₁ = 5 and n₂ = 3.

Thus, N = p₁⁵p₂³ = 2⁵3³ = 864, which is the smallest number so far.

d) We can also write 24 as 4(3)(2). Then, we get T = (n₁+1)(n₂+1)(n₃+1) = 4(3)(2) with

n₁ + 1 = 4, n₂ + 1 = 3, and n₃ + 1 = 2, giving us n₁ = 3, n₂ = 2, and n₃ = 1. So,

e) We can even write 24 as 2(2)(2)(3). Then, we get T = (n₁+1)(n₂+1)(n₃+1)(n₄+1) = 2(2)(2)(3)

with n₁ + 1 = 2, n₂ + 1 = 2, n₃ + 1 = 2, and n₄ + 1 = 3, giving us n₁ = 1, n₂ = 1, n₃ = 1,

and n₄ = 2. So, N = 2¹3¹5¹7² = 1,470, but it is silly to put the largest exponent 2 on the largest

prime 7. Putting the exponent 2 on the smallest prime 2, we get N = 2²3¹5¹7¹ = 420, but this,

We have examined all the possible ways of writing 24 as a product of two, three, or four primes, so

[1] Beiler, Albert H., Recreations in the Theory of Numbers: The Queen of Mathematics Entertains,

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