Count them without listing them — the exponents already know the answer.
For 360
How many divisors does it have?
360 = 2³ × 3² × 5. Add one to every exponent, then multiply: (3+1)(2+1)(1+1) = 24. The +1 exists because you may use each prime zero times. A forgets that 360 counts as a divisor of itself.
Every divisor has a partner: the number you multiply it by to reach the total.
| Pair | Product |
|---|---|
| 1 × 36 | 36 |
| 2 × 18 | 36 |
| 3 × 12 | 36 |
| 4 × 9 | 36 |
| 6 × 6 | the middle pair — one number |
Walk upward from 1 and stop at the square root. Everything above √36 = 6 is already written down as somebody's partner, so there is nothing left to find.
Factor it, add one to every exponent, multiply.
360 = 2³ · 3² · 5¹
the prime factorisation
(3 + 1)(2 + 1)(1 + 1)
each exponent, plus one
4 × 3 × 2 = 24
divisors, counted
The 5 has an invisible exponent of 1, so it contributes a factor of 2, not nothing. Forgetting it halves your answer — and 12 will be one of the choices.
Building a divisor is choosing how many of each prime to take — and zero is allowed.
2s to take
4
×
3s to take
3
×
5s to take
2
| Prime | Choices of exponent | How many |
|---|---|---|
| 2 | 0, 1, 2, 3 | 4 |
| 3 | 0, 1, 2 | 3 |
| 5 | 0, 1 | 2 |
Taking none of everything gives the divisor 1; taking all of everything gives 360. It is the counting principle, applied to exponents.
Divisors pair up, which would always give an even total — unless one divisor is its own partner. That happens only at √n, and only when √n is a whole number. So odd divisor count and perfect square mean the same thing.
Add up the powers of each prime separately, then multiply the totals.
(1 + 2 + 4 + 8)
powers of 2, up to 2³ → 15
(1 + 3 + 9)
powers of 3 → 13
(1 + 5)
powers of 5 → 6
15 × 13 × 6 = 1170
the sum of all 24
Multiplying those three brackets out produces every divisor exactly once — one term per combination of exponents. That is the same reason the count multiplied.
Odd, even, multiples of k, perfect squares — each is the same exponent count with the choices restricted.
| Count | Restriction | For 200 = 2³ · 5² |
|---|---|---|
| all divisors | none | (3 + 1)(2 + 1) = 12 |
| odd divisors | take no 2s at all | divisors of 5²: 1 × 3 = 3 |
| even divisors | all minus odd | 12 − 3 = 9 |
| multiples of 10 | 10 × a divisor of 200 ÷ 10 = 20 | 20 = 2² · 5: 3 × 2 = 6 |
How many divisors of 2160 = 2⁴ · 3³ · 5 are perfect squares?
a square divisor uses even exponents only
0 counts as even
2: 0, 2, 4 → 3 · 3: 0, 2 → 2 · 5: 0 → 1
the allowed choices
3 × 2 × 1 = 6
1, 4, 16, 9, 36, 144
For “odd divisors” the 2 gets exactly one choice (exponent 0), not zero choices. For “multiples of 10”, divide out the 10 first and count the divisors of what is left.
The number is a power: how many divisors does 12⁵ have?
12⁵ = (2² · 3)⁵ = 2¹⁰ · 3⁵
factor the base, multiply the exponents
(10 + 1)(5 + 1) = 66
then the usual rule
What is different: we never compute 12⁵. Counting the 6 divisors of 12 and multiplying by 5 gives 30, which is wrong — the exponents grow, not the count.
A different “divisor”: 84, counted three ways.
84 = 2² · 3 · 7 → 3 × 2 × 2 = 12
all positive divisors
other than 1 and 84: 12 − 2 = 10
drop the two ends
integer divisors: 2 × 12 = 24
each has a negative twin
What is different: the formula counts positive divisors including 1 and the number. Read which ones the question wants, then adjust.
Only one prime, or none: 1024 and 1.
1024 = 2¹⁰ → 10 + 1 = 11
one bracket: 1, 2, 4, …, 1024
1 has 1 divisor
no primes, so the product is empty
What is different: with a single prime there is nothing to multiply. 11 is odd, and 1024 = 32², as the perfect-square rule says.
What is the smallest positive integer with exactly 8 divisors?
8 = 8, or 4 × 2, or 2 × 2 × 2
each factor is an exponent + 1
p⁷, or p³q, or pqr
the three possible shapes
2⁷ = 128 · 2³ × 3 = 24 · 2 × 3 × 5 = 30
biggest exponent on the smallest prime
smallest: 24
1, 2, 3, 4, 6, 8, 12, 24 ✓
The same backwards step finds a missing exponent: if 2ᵏ · 5 has 12 divisors, then (k + 1)(1 + 1) = 12, so k + 1 = 6 and k = 5. And exactly 3 divisors always means the square of a prime, p², with divisors 1, p, p².
Trying only p⁷ gives 128, and 3³ × 2 = 54 puts the big exponent on the wrong prime. List each way of splitting the count, build the smallest number for each, and compare.
Divisor items ask for the count, the count of a kind (odd, even, square, multiples of k), which number has an odd count, or the smallest number with a given count. Factor first, then add one to each exponent you are allowed to vary — no listing beyond about 50.
Small enough to list, so the formula has to agree with the list.
The problem
How many divisors does 720 have?
720 = 2⁴ × 3² × 5, so (4+1)(2+1)(1+1) = 30. A is the answer for 360 — doubling a number does not double its divisor count, it only lifts one exponent by one.
