Connect Squares, Cubes, and Divisor Counts
连接平方数、立方数与约数个数
Use prime exponents to see when a number is a square, a cube, or both; then connect those exponent patterns to odd divisor counts and to numbers with exactly three divisors.
Meet squares, cubes, and their overlap
A square uses two equal factors. A cube uses three. Some positive integers have both forms.
Build with root 4
Checkpoint for the fixed numbers 4, 64 and 8
These questions keep the same numbers when you change the explorer.
Need a hint? Start here
A square uses two equal factors; a cube uses three. Test both forms separately.
4²=16,4³=64;64=8²=4³;8=2³ but lies between 2² and 3². 16;64;both;cube only.
Read square and cube rules from prime exponents
From this mission onward, all numbers are positive whole numbers and all divisors are positive. (The first explorer also allowed zero.) Prime factorization turns a large number into an exponent ledger. The exponent pattern determines the perfect-power category.
A number has exactly one distinct prime factor, with exponent 5. What is the least number of extra copies of that prime needed to make it both square and cube?
Your prediction is checked with this mission. Try it before reading the worked steps below.
The three exponent tests
| Property | Required exponent pattern |
|---|---|
| Perfect square | Every exponent is a multiple of 2 |
| Perfect cube | Every exponent is a multiple of 3 |
| Both | Every exponent is a multiple of lcm(2,3)=6 |
Exponent ledger
Exponent checkpoint
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Check every exponent for divisibility by 2 and by 3. Both tests require multiples of their LCM.
Their LCM is 6. The prediction needs one copy, changing exponent 5 to 6. Prediction 1. Checkpoint:6;square only;cube only;root 6.
Worked example 5: numbers below 1,000 that are both square and cube
The worked example asks about positive whole numbers below 1,000. Its solution converts the two power conditions into one sixth-power condition.
Turn two conditions into one
A square has prime exponents divisible by 2. A cube has prime exponents divisible by 3. Therefore a number with both properties is a sixth power:
| k | k⁶ | Below 1,000? |
|---|---|---|
| 1 | 1 | Yes |
| 2 | 64 | Yes |
| 3 | 729 | Yes |
| 4 | 4,096 | No |
original-example checkpoint
Need a hint? Start here
Generate positive sixth powers in order. Stop at the first value reaching the bound.
1⁶=1,2⁶=64,3⁶=729,4⁶=4096. Count3;64;729;4096.
Count common square–cube numbers below any bound
Instead of listing every square and every cube, generate the sixth powers directly.
Sixth-power boundary laboratory
Why this is complete
The sixth powers increase with the root:
Once one sixth power reaches the bound, every later one is too large.
Watch the endpoint
“Below 1,000” means:
It does not include a value equal to the bound.
Boundary checkpoint
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Pay attention to “below”: a value equal to the bound is excluded.
Below 10000:1,64,729,4096. Below65:1,64. 4;4096;2;64.
See why squares have an odd number of divisors
The lamp investigation depends on factor pairs. Every ordinary pair contributes two divisors; a square root pairs with itself.
The factor-pair mechanism
For a nonsquare, every divisor has a different partner:
That gives an even number of divisors.
For a square, the square root creates a self-pair:
That one unpaired root makes the divisor count odd.
Factor-pair explorer
Divisor-parity checkpoint
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Pair divisors around the root. A self-pair contributes just one divisor.
36 has four two-divisor pairs and self-pair6×6;30 has four two-divisor pairs. 9;8;odd;yes.
Exactly three divisors means the square of a prime
The practice problem asks for all two-digit natural numbers with exactly three positive divisors.
Why the form must be p²
A number with exactly three divisors must have the list:
The middle divisor p must be prime. If p were composite, it would contribute more divisors.
For a two-digit p², the prime root must be a one-digit prime whose square has two digits.
Two-digit search
Exactly-three-divisors range explorer
Both endpoints are included.
original-exercise checkpoint
Need a hint? Start here
Exactly three divisors leaves only 1, a prime p, and p squared.
Single-digit prime roots are2,3,5,7;only5²=25 and7²=49 have two digits. Two answers;25;49;sum74.
Audit one number from every viewpoint
Prime factorization, perfect-power status, and divisor count should tell one consistent story.
Divisor-structure laboratory
Example: 64
- 6 is even, so 64 is a square.
- 6 is divisible by 3, so 64 is a cube.
- τ(64)=6+1=7, an odd divisor count.
Example: 72
- 3 is not even, so not a square.
- 2 is not divisible by 3, so not a cube.
- τ(72)=(3+1)(2+1)=12.
Structure checkpoint
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For an exponent e, a divisor can use 0 through e copies: e+1 choices. Multiply independent choices.
64=2⁶ and729=3⁶ each have7 divisors;72=2³×3² is neither square nor cube. 7;7;neither;yes.
Search complete property families
Choose a property and an upper bound. The studio enumerates every positive integer in the range and reports the full family.
Equivalence check
For every positive bound, the two searches return the same family.
Special subfamily
This family sits inside the perfect squares but contains only squares of primes.
Search-studio checkpoint
Need a hint? Start here
Generate roots for squares, cubes and prime squares. Include 1 when positive values are allowed.
Squares:1²–10²;cubes:1³–4³;both:1,64;prime squares:4,9,25,49. 10;4;2;4.
Squares, cubes, and divisors workshop
Correct all eight answers to complete the workshop.
Optional reflection — not automatically graded
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Choose a method before calculating. Use the earlier local hint for that method, then verify the root, all stated conditions and the endpoints.
The least positive common exponent is6; two-digit prime squares are25 and49. Both;6;3;4096;9;2;7;square.
Exit ticket
Complete all ten missions, including all five exit questions, to earn your certificate.
Fresh challenge: use the methods from this lesson on these new values. Check your calculations by squaring, listing a short cycle, or verifying every condition.
Square–Cube & Divisor-Count Architect
This certifies that a determined mathematician can connect prime-exponent patterns, sixth powers, odd divisor counts, and prime squares with exactly three divisors.
Lesson 20.4 • Chapter 20
Need a hint? Start here
Choose a method before calculating. Use the earlier local hint for that method, then verify the root, all stated conditions and the endpoints.
50=2×5² has(1+1)(2+1)=6 divisors. Below10000 are1,64,729,4096. 6;729;25;6;four.