Balance Prime Exponents to Make a Perfect Square
平衡质因数的指数，使乘积成为完全平方数
Pair every prime factor, repair odd exponents, find the least multiplier, describe every possible multiplier, and use the method in a real age story.
Pair every prime copy
A positive whole number is a perfect square when every prime factor can be paired. We use positive whole-number starting values and multipliers throughout this lesson. The number 1 has no prime factors and is already a square; zero is outside these factorization problems. An odd leftover copy prevents the number from being a square.
Try a fresh number: 98=2×7². Predict its least positive square-completion multiplier.
Your prediction is checked with this mission. Try it before reading the worked steps below.
Why exponents matter
Suppose
Then N is a perfect square exactly when a, b, and c are all even. Each exponent counts how many copies of that prime appear.
Odd exponent → one copy is left unpaired.
Compare two numbers
72
The exponent 3 is odd. One 2 is unpaired.
144
All exponents are even, so 144 = 12².
Pairing checkpoint
Need a hint? Start here
Write the prime copies in pairs. Only an unmatched copy needs another partner.
For the prediction, 98=2×7² needs one 2. For 72=2³×3², add one 2 to make 2⁴×3². Prediction 2. Checkpoint: 3; odd; 2; 144.
Read an exponent ledger
Adjust the exponents and watch the page decide whether the number is already a square and which prime copies are missing.
Checkpoint for the fixed exponents 5, 2, 3, 1
Set the exponents to 2⁵ × 3² × 5³ × 7¹, then answer.
Need a hint? Start here
For each odd exponent, add one copy of its prime. Halve the repaired exponents to find the root.
The odd columns are 2,5,7, so multiply by 70. The new exponents are 6,2,4,2. Three odd exponents; 70; yes; root 2³×3×5²×7=4200.
Worked example 2: complete 46,035 to a square
This worked example asks for the least positive whole number a such that 46,035 × a is a perfect square.
Factor the given number
Every displayed exponent is odd.
Supply one matching copy
Then all exponents become even.
Select the primes that belong in the least multiplier
original-example checkpoint
Need a hint? Start here
The missing partners belong to the primes with odd exponents. Multiply those primes once each.
3³,5¹,11¹,31¹ all have odd exponents. Multiply one copy of each. Select 3,5,11,31. Answers: 5115; 235469025; 15345; exponent 4.
Guided Practice 2: complete 203,500
Work through this guided repair of 203,500. Explain each step, then use the workshop and exit questions to test the method independently.
Prime factorization
The exponent of 2 is already even. The exponents of 5, 11, and 37 are odd.
Repair preview
Every exponent is now even.
Practice checkpoint
Need a hint? Start here
The exponent of 2 is already even. Check the other three prime columns.
Only 5,11,37 need partners: their product is 2035. 2035; 414122500; root 20350; three different primes.
Use a general least-square-multiplier laboratory
Enter any positive whole number. The laboratory factors it, marks odd exponents, finds the least multiplier, and verifies the resulting square.
Checkpoint for the fixed starting number 1,080
Load 1,080, then answer.
Need a hint? Start here
Factor the input, repair odd exponents, then square your proposed root to verify the product.
1080=2³×3³×5. Multiply by 2×3×5=30. 30; 32400; root 180; three odd exponents.
Describe every multiplier, not only the least one
Once the least repair multiplier s is known, every positive whole-number multiplier that works has the form s × k², where k = 1, 2, 3, … .
Why the pattern works
If N × s is a square, then multiplying by another square keeps it a square:
Conversely, after the required odd exponents are repaired, every additional exponent must be even. So every extra factor is a square.
Example with 72
First few values:
Multiplier-family explorer
Family checkpoint
Need a hint? Start here
After the least repair, extra prime copies must come in pairs. That extra factor is a square.
Use x=2k². At k=1,2,3 the multipliers are 2,8,18. Least 2; form 2k²; third 18; product 1296.
original Exercise 9: count every valid number below 2,008
The worked example asks how many positive whole numbers n strictly below 2,008 make 72n a perfect square.
Turn the condition into a family
The bound is:
Because 31² = 961 but 32² = 1024, the possible k values are 1 through 31.
Complete runtime list
Count a family below any bound
original-exercise checkpoint
Need a hint? Start here
Substitute the multiplier family into the strict inequality. Compare the last allowed root with the next one.
31²=961<1004, but 32²=1024>1004. Largest k 31; count 31; largest n 1922; 2048 is excluded by the bound.
Use exponent balance in stories and transfer problems
Prime-exponent parity can recover a real-world role, but the story’s range may be needed to select one practical value.
Grandson × grandfather = 1,512
Assume the father has a whole-number age from 18 through 80 years. Use this range as part of the problem.
For grandson × father × grandfather to be a square, the father’s age must have the form:
Practice Question 2
If 32x is a perfect square, find the least positive x.
One more copy of 2 repairs the odd exponent.
Practice Question 3
Factorial means multiply every positive whole number up to n: for example, 4! = 1 × 2 × 3 × 4 = 24. Find the least whole number n > 1 such that:
is a perfect square. Test the smallest allowed n values in order.
Transfer checkpoint
Need a hint? Start here
Use the stated age range after exponent repair. For the factorial task, test the smallest allowed whole numbers in order.
1512=2³×3³×7 needs 42. Father ages 42k² start 42,168,…; only 42 is in 18–80. For factorials, 2!+3=5 and 3!+3=9. 42; 2; 3; 9.
Prime-exponent workshop
Correct all eight answers to complete the workshop. All variables and multipliers in these questions are positive whole numbers; in a family formula, k = 1, 2, 3, … .
Optional reflection — not automatically graded
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.
All working multipliers are the least multiplier times k² for positive whole-number k. Answers: even; 2; 5115; 2035; 30; 2k²; 31; 2.
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.
Prime-Exponent Square Architect
This certifies that a determined mathematician can pair prime factors, repair odd exponents, find least and general multipliers, and verify completed perfect squares.
Lesson 20.2 • 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.
Only the exponent of 2 is odd in either number. 2; 100; root 10; 2; form 2k² with k=1,2,3,… .
Learning notes and instructional additions
original Example 2 uses 46,035 = 3³ × 5 × 11 × 31 and gives the least multiplier 5,115.Guided Practice 2 asks the corresponding question for 203,500.The general factorization laboratory, proof that all working multipliers have the form s × k², exhaustive family enumeration, original-context note, workshop, and exit ticket are instructional additions.