Count and Sum Divisors from Prime Exponents
用质因数指数数出约数，并求约数和
Here, divisors always mean positive whole-number divisors, including 1 and the number itself. Turn a standard prime factorization into a complete divisor inventory. Count every divisor, filter divisors by size, add all divisors efficiently, and work backward from a required divisor count.
Every divisor is an exponent choice
For 72 = 2³ × 3², a divisor may use 0, 1, 2, or 3 copies of 2 and 0, 1, or 2 copies of 3. Every pair of choices creates one divisor.
The complete rule
where
Exponent-choice rectangle
Click cells to uncover the divisor made by that exponent pair.
Read the rectangle
Need a hint? Start here
Choosing zero copies contributes 1. For example, 2² × 3¹ makes divisor 12.
How can I check my reasoning?
There are four choices for i and three for j. 4; 3; 12; multiply. Each of the 4 × 3 cells gives a different divisor of 72.
Derive the divisor-count formula
If a prime appears with exponent a, a divisor can use 0 through a copies of that prime—exactly a + 1 choices.
General form
Use the standard factorization of N greater than 1: p₁, p₂, … are different primes, and their exponents are positive whole numbers.
τ(N) is a compact name for the number of positive divisors of N.
Apply it to 360
Common mistake
This counts prime copies, not divisors. Divisors come from combinations of exponent choices.
Complete the 360 count
Need a hint? Start here
List exponent choices starting at zero, then multiply the numbers of choices.
How can I check my reasoning?
For 360=2³×3²×5, the choice counts are 4,3,2. 4 × 3 × 2 = 24 positive divisors, including 1 and 360.
Filter divisors by digit length
original Exercise 1 asks for each standard prime factorization and the number of two-digit divisors. Generate all divisors first, then keep only those from 10 through 99.
original Exercise 1 — first two numbers
Prime-factorize 146 and 255, then identify their two-digit positive divisors. Enter only the factorization, for example 2^3 * 3^2: use increasing prime bases and combine repeated copies. Do not enter an equals sign.
| Number | Standard factorization | Two-digit divisors (commas between values; any order) | Count |
|---|---|---|---|
| 146 | |||
| 255 |
Need a hint? Start here
First list all divisors; retain only values from 10 through 99.
How can I check my reasoning?
Use increasing prime bases in your factorizations; comma-separated divisor lists may be in any order. 146=2×73 gives 73 only. 255=3×5×17 gives 15,17,51,85. Counts: 1 and 4.
Generate, count, sum, and filter every divisor
Finish original Exercise 1 for 360 and 400, then use the general laboratory for any whole number from 2 through 1,000,000.
original values
| N | τ(N) | Two-digit divisor count |
|---|---|---|
| 360 | ||
| 400 |
Divisor laboratory
Choose a preset or enter a number, then generate its full divisor inventory. τ(N) counts all positive divisors and σ(N) sums them all; the range affects only the highlighted subset and its count.
Need a hint? Start here
Count every exponent combination once, then apply the digit filter.
How can I check my reasoning?
360 has exponent choices 4×3×2; 400=2⁴×5² has 5×3. 360: 24 total, 13 two-digit. 400: 15 total, 7 two-digit (10,16,20,25,40,50,80).
Sum divisors by multiplying mini-sums
The same exponent choices that count divisors can add them. Distributive expansion creates every divisor exactly once.
Build the formula with 12
General divisor-sum formula
σ(N) is a compact name for the sum of all positive divisors.
Apply the formula to 72
Divisor-sum laboratory
Need a hint? Start here
Expand a product of sums: each selected term makes one divisor.
How can I check my reasoning?
For 72, add 1+2+4+8 and 1+3+9 separately. 15 × 13 = 195. Expanding selects one power from each group, giving each divisor exactly once.
Solve the practice problem for 165
original Exercise 3 asks both how many divisors 165 has and what all those divisors add to.
original Exercise 3
Each exponent is 1, so each prime may be used either 0 times or 1 time.
Need a hint? Start here
165 ends in 5; after dividing, factor what remains.
How can I check my reasoning?
Each prime can be absent or present: 2×2×2 choices. Eight divisors: 1,3,5,11,15,33,55,165. Their sum is (1+3)(1+5)(1+11)=288.
Work backward from a divisor count
original Exercise 4 asks for the smallest number with exactly nine positive divisors. Work backward from the product of the “exponent plus one” factors.
How can τ(N) equal 9?
This gives one exponent:
This gives two exponents:
Use the smallest primes
Shape p⁸
Shape p²q²
Complete original Exercise 4
Divisor-count architect
Need a hint? Start here
Nine is either 9 or 3 × 3. Subtract 1 to recover possible exponents.
How can I check my reasoning?
Only 9 and 3×3 are possible. Subtract 1 from each factor. Exponent patterns: 8 and 2,2. The smallest candidates are 2⁸=256 and 2²×3²=36, so 36 is smallest overall.
Protect special cases and count many products efficiently
Two original exercises reveal different dangers: a hidden special case can split the answer, while a large product table can collapse by the distributive property.
original Exercise 2 — a necessary case split
The worked example says: “A positive whole number a has exactly two divisors. How many divisors does 3a have?” Here 3a means 3 × a. Therefore a is prime.
original Exercise 13 — sum all prime–composite products
Choose one prime and one composite from the lists below, multiply them, and add the results for all 4 × 4 = 16 pairs. Count each pair once. Among the one-digit positive integers:
State the complete results
Need a hint? Start here
Separate a = 3 from the other primes before using the counting formula.
How can I check my reasoning?
For the product table, multiply the sum of the four primes by the sum of the four composites. If a=3, 3a=9 has 3 divisors. Otherwise 3a has 4. The sums are 17 and 27, so the 16 products total 459.
Divisor-count and divisor-sum workshop
Correct all eight answers to complete the workshop.
Need a hint? Start here
Choose the strategy first: factor, apply the conditions, and multiply back. Revisit the relevant local hint above.
How can I check my reasoning?
Use the exponent-choice formula for counts and mini-sums for sums. Answers: 12; 4; 28; 15; 36; 288; 3; 459.
Objective exit ticket
Complete all ten missions, including 5 out of 5 on this exit ticket, to earn the certificate.
Prime-Exponent Divisor Architect
This certifies that
can generate divisors from prime exponents, count and sum them, filter them by size, and work backward from a divisor count.
Need a hint? Start here
Work without the laboratories first. Explain which prime copies or factor pairs your answer uses, then verify the result.
How can I check my reasoning?
20=2²×5, so its divisor sum is (1+2+4)(1+5). Answers: 15; 42; 1; 36; 4.
Teaching notes
The original asks for the prime factorizations and two-digit-divisor counts of 146, 255, 360, and 400; the divisor count and divisor sum of 165; the smallest number with nine divisors; and the sum of all products formed from a one-digit prime and a one-digit composite.Exercise 2 is presented with an explicit case split because the original wording allows a = 3 as well as other primes.