🛍️
Most identical bags
Every bag must receive the same contents, with nothing left.
Use GCDMaximize Common Factors and Share Quantities Equally
把和分成若干数时最大化公约数；平均分组与等长截段
Use divisors, positivity bounds, and concrete constructions to maximize a shared GCD. Then turn the same idea into the greatest number of identical groups, equal bags, teams, and longest no-waste cuts.
A maximum needs two parts: an upper bound and a construction that reaches it.
Answers autosave in this browser.
Mission 1
Not completeChoose the right tool before calculating
GCD and LCM can both appear in grouping stories. The deciding question is whether you seek the greatest shared unit or the first shared multiple.
✂️
Longest equal pieces
Every original length must be cut exactly, with no remainder.
Use GCD🔁
First shared return
Several repeating cycles start together and must meet again for the first time after that start.
Use LCM👥
Least possible total
The unknown total must be divisible by several allowed group sizes.
Use LCMMethod checkpoint
Need a hint? Start here
Identical groups and no-waste cuts use divisors; a least compatible total uses multiples.
Check why it works
No-waste identical bags and equal cuts use common divisors; schedules begin together. GCD; GCD; LCM; LCM.
Mission 2
Not completeProve the fixed-sum bound
Suppose k positive natural numbers have sum S and common divisor g. Two restrictions immediately control how large g may be.
Restriction 1: divide the total
S = g(c₁ + c₂ + ··· + cₖ)
Because every number is a multiple of g, their sum is also a multiple of g.
So g must divide S.
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Restriction 2: fit all positive numbers
each number ≥ g
kg ≤ S
There must be enough total value for all k positive numbers to contain at least one copy of g.
This lesson uses positive whole numbers: 1, 2, 3, … Repeated terms are allowed. Each cᵢ is the positive whole-number coefficient obtained by dividing that term by g.
Candidate tester
Bound checkpoint
Need a hint? Start here
Every positive term contains at least one copy of the proposed common divisor.
Check why it works
Thus g divides S and kg cannot exceed S. Divide the total; kg≤S; no, four positive multiples of 1111 cannot sum to 1111.
Mission 3
Not completeSolve Worked example 5: four numbers total 1111
The worked example asks for the greatest possible common divisor of four natural numbers whose sum is 1111.
Filter the divisors of 1111
1111 = 11 × 101
The positive divisors are:
1111011111
The four-number bound is 4g ≤ 1111, so g ≤ 277.75. The divisor 1111 is too large.
The largest divisor that also fits the positivity bound is 101.
Construct four numbers that reach the bound
11 = 1 + 2 + 3 + 5
101202303505
101 + 202 + 303 + 505 = 1111
gcd(101,202,303,505) = 101
The coefficient GCD is gcd(1,2,3,5)=1, so the four numbers' GCD is exactly 101—not a larger multiple.
original checkpoint
Need a hint? Start here
List divisors of the total, apply the size bound, then construct a matching example.
Check why it works
101 is the largest allowed divisor. The coefficients 1,2,3,5 have GCD 1. Maximum GCD 101; coefficient sum 11; constructed sum 1111.
Mission 4
Not completeUse a general fixed-sum maximum-GCD laboratory
The maximum is the greatest divisor of the total that does not exceed S ÷ k. The laboratory also constructs positive numbers whose GCD really equals that value.
Use a total up to 1,000,000 and 2–1,000 positive terms. Repeated terms are allowed.
Independent checkpoint: 1000 split into 6 positive numbers
Use five copies of the maximum GCD and one final term that supplies the remaining total. The boxes ask about this specific construction; other optimal constructions may also exist.
Need a hint? Start here
Use k−1 copies of the proposed GCD, then use the remaining total as the last term.
Check why it works
The largest divisor no greater than 166⅔ is 125. Use five copies, then subtract their sum from 1000. Maximum GCD 125; repeated term 125; final term 375.
Mission 5
Not completeComplete Guided Practice 5: write 2001 as 25 numbers
The 25 natural numbers may be equal or different. We still need both a divisor bound and a construction.
Find the largest divisor below the bound
2001 = 3 × 23 × 29
25g ≤ 2001 ⇒ g ≤ 80.04
13232969876672001
Construct 25 positive numbers
2001 ÷ 69 = 29
Make the 25 coefficient values:
1,1,…,1 (24 copies), 5
24 × 69 + 5 × 69 = 2001
69 × 24345
The coefficient list contains 1, so the GCD is exactly 69.
Guided-practice checkpoint
Need a hint? Start here
The coefficient sum is 2001 divided by the candidate GCD. Keep all 25 coefficients positive.
Check why it works
69 is the largest allowed divisor. 2001/69=29 coefficient units for 25 terms. Maximum 69; 24 copies of 69; final number 5×69=345.
Mission 6
Not completeShare several quantities equally
When every student must receive the same amount of every item and everything is distributed, the greatest possible number of students is the GCD of all totals.
Use up to ten whole-number totals from 1 through 1,000,000.
For these totals, the only common divisors are 1 and 43. If the class must contain more than one student, that condition alone selects 43. The greatest-number-of-students version below gives the same result.
Clarification: The printed exercise asks what each student receives but does not explicitly say to use the greatest possible number of students. Without that added condition, several group counts are possible. The GCD solution shown here uses the chapter-intended interpretation: share among as many students as possible.
301 notebooks, 215 pencils, 86 erasers
30143 × 77 each
21543 × 55 each
8643 × 22 each
gcd(301,215,86) = 43
So 43 students can each receive 7 notebooks, 5 pencils, and 2 erasers.
Equal-distribution laboratory
original checkpoint
Need a hint? Start here
A recipient count must divide every item total. Divide each total by the chosen count.
Check why it works
GCD(301,215,86)=43. Divide each kind by 43 separately. 43 students; 7 notebooks,5 pencils,2 erasers per student.
Mission 7
Not completePack the greatest number of identical bags
Test 19 asks for the maximum number of bags when 252 red, 396 blue, and 468 yellow balls must be used completely and every bag must have identical contents.
Use the GCD as the bag count
252 = 36 × 7
396 = 36 × 11
468 = 36 × 13
Maximum bags = 36
red7blue 11 · yellow 13
red7blue 11 · yellow 13
red7blue 11 · yellow 13
red7blue 11 · yellow 13
…36identical bags
all ballsusedno remainder
Candidate bag-count tester
Candidate bags36
Maximum possible36
Bagging checkpoint
Find the maximum for the fixed color totals, even if you try a smaller valid bag count in the tester.
Need a hint? Start here
A valid bag count divides all three color totals. The greatest such count is their GCD.
Check why it works
GCD(252,396,468)=36. Smaller divisors give valid but fewer bags. 36 bags; 7 red,11 blue,13 yellow per bag.
Mission 8
Not completeCut equal pieces, form teams, and avoid a GCD–LCM mix-up
The same GCD idea controls longest no-waste cuts and the greatest number of identical teams. A least unknown total, however, is an LCM question.
Equal-cut laboratory
Measure every original length in the same unit. Enter positive whole-number lengths. Ignore material lost in cutting; use every length completely. The strips count pieces and are not drawn to a common scale.
Two more original-style decisions
120 boys
140 girlsgcd(120,140)=2020 teams
140 girlsgcd(120,140)=2020 teams
Use all 120 boys and 140 girls. Every team must have the same number of boys as every other team, and the same number of girls as every other team. Each of the 20 teams has 6 boys and 7 girls.
60–100 peoplemust be groupable with no remainder either into groups of 8 or into groups of 12 (separate arrangements)LCM
Here the unknown is the total number of people, so it must be a common multiple. The interval includes both 60 and 100. Its least valid total is 72.
Cutting and grouping checkpoint
Need a hint? Start here
Keep units consistent; divide each wire length by the common cut length and add the piece counts.
Check why it works
The cut length is 60, making 2+3+5=10 pieces. Teams must have matching boy and girl counts across teams. 60 length units; 10 pieces; 20 teams; least compatible total in 60–100 is 72.
Mission 9
Not completeIndependent equal-sharing workshop
Correct all eight questions to complete the workshop.
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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 maximum needs both an upper bound and an example reaching it. Answers: 101; 69; 43; 7; 36; 60; 10; 20.
Mission 10
Not completeExit ticket
Complete all ten missions and answer all five exit items correctly to earn your certificate.
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GCD
Equal-Share GCD Architect
This certifies that a determined mathematician can prove fixed-sum GCD maxima, build constructions that attain them, make the greatest number of identical groups, and find longest equal cuts without waste.
Lesson 19.5 • 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
95 is largest; seven 95s and 285 give an example. For the cuts, 96/48+144/48=5 pieces. Answers: 95; 42; 2; 5; 72.
Learning notes and instructional additions
Exercise 4 does not explicitly say to maximize the number of students, so the page visibly labels that chapter-intended assumption instead of silently adding it.The lesson explicitly uses positive natural numbers in fixed-sum problems because the original maximum argument and constructions require positive terms.The general laboratories, candidate diagnostics, and extra transfer examples are instructional scaffolds derived from the original methods.