GCD takes the minimum
If the exponents of prime 2 are 2 and 5, the GCD contains 2².
Keep only the prime copies shared by both numbers.
Recover Numbers from GCD, LCM, and Product
从最大公约数、最小公倍数和乘积反求两数
Use prime-exponent balance, the two-number product identity, coprime reduced factors, sums, products, and divisor counts to reconstruct hidden numbers—and verify that every recovered candidate really has the stated GCD and LCM.
For two positive integers: GCD × LCM = first number × second number.
Answers autosave in this browser.
Mission 1
Not completeBalance the prime exponents
For each prime, the GCD keeps the smaller exponent and the LCM keeps the larger exponent. Smaller plus larger equals the two original exponents added together.
LCM takes the maximum
For the same exponents, the LCM contains 2⁵.
Collect every prime copy needed by either number.
The exponents rebalance
For every prime:
min(r,s) + max(r,s) = r + s
Multiplying across all primes gives the identity.
Prime-exponent balance laboratory
Number A
Number B
Identity checkpoint
Use the fixed exponents 2 and 5 below, even if you change the laboratory. Exponent zero means no copies of that prime, so a number can equal 1. The product identity is not a general three-number rule: gcd(2,4,8)×lcm(2,4,8)=16, while 2×4×8=64.
Need a hint? Start here
For each prime, smaller exponent plus larger exponent equals the two starting exponents added.
Check why it works
The fixed exponent pair is 2 and 5. Two positive integers; GCD exponent 2; LCM exponent 5. The exponents add to 7 on both sides.
Mission 2
Not completeProve the worked example identity with 36 and 54
Worked example 4 uses prime exponents to prove the product identity for two numbers.
Worked example 4
Align the factorizations
36 = 2² × 3²
54 = 2¹ × 3³
| Prime | 36 | 54 | GCD: min | LCM: max |
|---|---|---|---|---|
| 2 | 2 | 1 | 1 | 2 |
| 3 | 2 | 3 | 2 | 3 |
Complete the balance
gcd(36,54) = 2 × 3² = 18
lcm(36,54) = 2² × 3³ = 108
GCD × LCM18 × 108
=
Original product36 × 54
1944 = 1944
Visible original correction. The problem statement and every calculation use b = 54, but one explanatory sentence on prints b = 45. That sentence is a typographical error; this lesson preserves the supported value 54 and identifies the correction openly.
original checkpoint
Answer for the fixed original example in this mission, even if you change the laboratory.
Need a hint? Start here
Multiply the GCD and LCM, then compare with the original product.
Check why it works
GCD 2×3²; LCM 2²×3³. 18; 108; common product 1944.
Mission 3
Not completeComplete Guided Practice 4
Apply the same exponent balance to 108 and 720, then verify both sides of the identity.
Guided Practice 4
Prime-exponent table
108 = 2² × 3³ × 5⁰
720 = 2⁴ × 3² × 5¹
| Prime | 108 | 720 | Minimum | Maximum |
|---|---|---|---|---|
| 2 | 2 | 4 | 2 | 4 |
| 3 | 3 | 2 | 2 | 3 |
| 5 | 0 | 1 | 0 | 1 |
Results
GCD = 2² × 3² = 36
LCM = 2⁴ × 3³ × 5 = 2160
36 × 2160 = 77760
108 × 720 = 77760
The two products agree prime by prime, not by coincidence.
Guided-practice checkpoint
Need a hint? Start here
Include a zero exponent for the missing factor 5 in 108.
Check why it works
GCD 2²×3²; LCM 2⁴×3³×5. 36; 2160; common product 77760.
Mission 4
Not completeRecover one missing number—and verify it
If one number a, the claimed GCD g, and the claimed LCM L are known, the identity suggests a candidate for the other number.
This calculator accepts positive integers up to 1,000,000,000 and keeps the intermediate product exact.
a × b = g × L
→
b = gL ÷ a
→
Check gcd(a,b)=g and lcm(a,b)=L
original Exercise 9
Given:
a = 24, gcd(a,b) = 4, lcm(a,b) = 168
b = 4 × 168 ÷ 24 = 28
gcd(24,28)4 ✓
lcm(24,28)168 ✓
The verification matters. The product formula can produce a number even when the claimed GCD and LCM data are inconsistent.
Missing-number laboratory
original checkpoint
Answer for the fixed original example in this mission, even if you change the laboratory.
Need a hint? Start here
Calculate the candidate, then verify its actual GCD and LCM.
Check why it works
4×168÷24=28; 24=4×6 and 28=4×7, with 6 and 7 coprime. Recovered b 28; actual GCD 4; actual LCM 168.
Mission 5
Not completeRecover every pair from a GCD and an LCM
Write the two numbers as a = gx and b = gy. The reduced factors must be coprime, and their product is L ÷ g.
Use positive GCD and LCM values up to 1,000,000.
Why the reduced factors are coprime
a = gx, b = gy
If x and y shared another factor, then a and b would share more than g. Therefore:
gcd(x,y) = 1
Using the product identity:
g²xy = gL ⟹ xy = L ÷ g
GCD–LCM pair explorer
All original and reduced factors are positive whole numbers. A reduced factor may equal 1, and the original numbers may be equal. Reversing a pair counts as the same unordered pair.
Study-case checkpoint: g = 6, L = 180
Need a hint? Start here
Divide both numbers by their GCD. The two remaining factors must be coprime.
Check why it works
Reduced pairs: (1,30),(2,15),(3,10),(5,6). Multiply each member by 6. Reduced product 30; four pairs (6,180),(12,90),(18,60),(30,36). The smallest difference belongs to (30,36), whose smaller member is 30.
Mission 6
Not completeAdd a fixed sum—and keep every valid outcome
Find every pair of positive whole numbers with sum 50 and GCD 5. Here “difference” means larger minus smaller. Determine all possible differences rather than assuming there is only one.
The search supports sums and GCDs up to 10,000.
Practice Question 3
Reduce by the GCD
a = 5x, b = 5y
x + y = 50 ÷ 5 = 10
Keep only coprime positive pairs:
x,y = 1,9
a,b = 5,45
Difference = 40
x,y = 3,7
a,b = 15,35
Difference = 20
The printed data do not determine one difference. A complete answer must preserve both valid outcomes.
Fixed-sum and GCD laboratory
original-test checkpoint
Need a hint? Start here
Divide the sum by the GCD, then list all coprime pairs with that sum.
Check why it works
Coprime positive pairs totaling 10 are (1,9) and (3,7). Original pairs (5,45) and (15,35). Larger difference 40; smaller difference 20; two possibilities.
Mission 7
Not completeRecover two numbers from their product and GCD
original Exercise 5 adds an important condition: both recovered numbers must be composite.
Use a product and a GCD no greater than 1,000,000.
original Exercise 5
ab = 5766, gcd(a,b) = 31
a = 31x, b = 31y, gcd(x,y)=1
xy = 5766 ÷ 31² = 6
x,y = 1,6
a,b = 31,186
Reject: 31 is primex,y = 2,3
a,b = 62,93
Both composite62 × 93 = 5766
Product-and-GCD laboratory
original checkpoint
Answer for the fixed original example in this mission, even if you change the laboratory.
Need a hint? Start here
Divide the product by the square of the GCD. Check the composite condition last.
Check why it works
Reduced product 6 gives (1,6) or (2,3). The first produces prime 31, violating the composite restriction. The valid numbers are 62 and 93, with sum 155.
Mission 8
Not completeCombine a GCD with divisor counts
original Exercise 10 says the only possible prime factors of A and B are 3 and 5, their GCD is 75, A has 12 positive divisors, and B has 10. Since 75=3×5² divides both, each must contain at least one 3 and two 5s.
Translate every clue into exponents
75 = 3¹ × 5²
Write:
A = 3ᵃ × 5ᵇ, B = 3ᶜ × 5ᵈ
min(a,c)=1, min(b,d)=2
| Number | Divisor count | Possible exponent pairs |
|---|---|---|
| A | (a+1)(b+1)=12 | (1,5), (2,3), (3,2), (5,1) |
| B | (c+1)(d+1)=10 | (1,4), (4,1) |
Try filtering the exponent pairs, then reveal the worked solution
- Both 5-exponents must be at least 2, so B cannot use (4,1). Therefore (c,d)=(1,4).
- Now min(b,4)=2, so A must have b=2.
- Among A's divisor-count pairs, only (a,b)=(3,2) works.
A = 3³ × 5² = 675
B = 3¹ × 5⁴ = 1875
A + B = 2550
original checkpoint
Answer for the fixed original example in this mission, even if you change the laboratory.
Need a hint? Start here
A divisor count is the product of one more than each exponent. Then apply the minimum-exponent clues.
Check why it works
B must use exponents (1,4); then A needs 5-exponent 2 and thus 3-exponent 3. a=3,b=2,c=1,d=4. A=675, B=1875; A+B=2550.
Mission 9
Not completeIndependent reconstruction workshop
Correct all eight questions to complete the workshop.
0 / 8
Need a hint? Start here
Choose the method before calculating. Revisit the local hint for that method, then verify every condition in the question.
Check why it works
A product identity is necessary but does not by itself verify the claimed GCD. Answers: 14; 210; 18; 28; 4; 40; 62; 2550.
Mission 10
Not completeExit ticket
Complete all ten missions and answer all five exit items correctly to earn your certificate.
0 / 5
gL
GCD–LCM Reconstruction Architect
This certifies that a determined mathematician can balance prime exponents, use the two-number product identity, reconstruct all valid pairs, and verify GCD, LCM, sum, product, compositeness, and divisor-count conditions.
Lesson 19.4 • Chapter 19
Optional reflection — not automatically graded
Need a hint? Start here
Choose the method before calculating. Revisit the local hint for that method, then verify every condition in the question.
Check why it works
For g=8,L=120, reduced product 15 gives (1,15) and (3,5). Answers: 16; 240; 60; 2; 2550.
Learning notes and instructional additions
The lesson visibly identifies the page's isolated “45” typo because the problem statement and all calculations support 54.The general recovery laboratories and inconsistent-data examples are instructional scaffolds derived from the original methods.