LCM: first repeat
Use the LCM for the next shared date, the first shared footprint position, or a simultaneous return.
Ask: “When or where do all cycles line up again?”
Synchronize Cycles, Spacing, and Consecutive Sums
同步周期、间距与连续数之和
Use least common multiples to locate the first repeated distance or date, count shared marks without double-counting, and verify when one total can be written in several consecutive-number ways.
The LCM finds the first positive place where repeating patterns line up.
Answers autosave in this browser.
Mission 1
Not completeDecide whether the story needs a GCD or an LCM
A greatest common divisor divides a fixed amount into the greatest equal units. A least common multiple identifies the first positive point shared by repeating patterns.
GCD: greatest common unit
Use the GCD for the longest equal cut, greatest number of identical groups, or largest unit fitting every amount.
Ask: “What greatest size divides every quantity?”
Then verify the story
An LCM gives a necessary repeat point, but a construction may still be needed—especially in consecutive-sum problems.
A common multiple is not enough; use the least positive one and check all conditions.
Watch two cycles meet
Markers appear every 6 and every 8 units. Their first positive shared marker is at 24.
Every 6Every 8Shared
Method checkpoint
Need a hint? Start here
Ask whether you need a greatest dividing unit or a least shared multiple.
Check why it works
For the repeat questions, the schedules begin together. LCM; GCD; LCM.
Mission 2
Not completeFind the first shared footprint position
In Worked example 3, a child takes 54-centimetre steps and the father takes 72-centimetre steps along the same circular path. Treat each footprint as a point on the path. Both start at the same point and finish one lap using only full steps, so the circumference is a multiple of both step lengths. Starting together, their footprints first coincide again after one least common multiple.
Worked example 3 — one repeat section
lcm(54, 72) = 216 cm
Within the positive positions up to and including 216 cm:
- the child leaves footprints at 54, 108, 162, and 216;
- the father leaves footprints at 72, 144, and 216;
- the footprint at 216 is shared.
For each repeat section, count positions after its start, up to and including its end. This prevents double-counting internal boundaries. Only the end of the full lap returns to the original starting place.
Footprint strip: 0 to 216 cm
Child • 54 cm
Father • 72 cm
Child onlyFather onlyShared at 216
original checkpoint
Need a hint? Start here
Find the first positive position that is a multiple of both step lengths.
Check why it works
216÷54=4; 216÷72=3. Count positions after the start through the endpoint. Repeat distance 216 cm; child count 4; father count 3.
Mission 3
Not completeCount distinct marks and recover the circular path
In every 216-centimetre repeat section, add the two footprint counts and subtract the shared endpoint once.
One section
4 + 3 − 1 = 6 distinct footprints
The snow shows 60 distinct footprint positions after both walkers complete one lap. Overlapping footprints count as one position. The final shared position is the starting point, so do not add another mark for the start. Therefore:
60 ÷ 6 = 10 repeat sections
216 × 10 = 2160 cm = 21.6 m
Build the loop from repeat sections
Distinct footprints60
Circumference2160 cm
In metres21.6 m
Repeat length216 cm
Finish the worked example
Set the section-count slider to 10 as well as entering your answers for the fixed 60-mark problem.
Need a hint? Start here
Add both counts, then subtract the shared mark once.
Check why it works
Each section has 4+3−1=6 distinct positions. Set the slider to 10. 60÷6=10 sections; circumference 2160 cm=21.6 m. Do not add the starting point again.
Mission 4
Not completeUse a general spacing-and-overlap laboratory
For step lengths a and b, one repeat section has length lcm(a,b). Count the marks from each walker and subtract the shared endpoint once.
Challenge checkpoint: 30 m, 45 m, and 20 distinct marks
Use the same circular-path model: both walkers finish at their common start using full steps. Length answers are in metres.
Need a hint? Start here
The observed marks must make a whole number of complete repeat sections.
Check why it works
Repeat 90 m; marks per section 3+2−1=4; sections 20÷4=5. 90 m; 4 marks; 450 m.
Mission 5
Not completeSynchronize repeating visits on a calendar
Guided Practice 3 says three people visit every 6, 8, and 9 days. If they meet on March 5, the next shared visit occurs one LCM later.
Choose a starting year from 1900 through 2099. Intervals must be whole numbers from 1 to 366 days; the shared interval may span at most 36,600 days. The result includes its year, so leap days are handled across year boundaries.
Guided Practice 3
lcm(6, 8, 9) = 72 days
The starting meeting is Day 0. Add 72 full days:
March 5Start togetherDay 0
March 31Move 26 days26 used
April 30Move 30 more days56 used
May 16Move the final 16 days72 used
Calendar synchronization laboratory
The checkpoint below always asks about the March 5 start and intervals 6,8,9 days, even if you explore another date.
original checkpoint
Need a hint? Start here
Treat the starting meeting as Day 0, then add the LCM in full days.
Check why it works
LCM(6,8,9)=72 days. March uses 26 days, April 30, leaving 16. 72 days; May; 16.
Mission 6
Not completeShift several cycles by the same remainder
Sometimes the shared condition is not “remainder 0.” If every grouping leaves the same remainder r, subtract r first, synchronize the remaining multiples, then add r back.
Use at most ten cycles, each no greater than 1,000,000,000. This laboratory supports a combined LCM up to 1,000,000,000.
Shifted-cycle laboratory
Enter a nonnegative remainder smaller than every cycle length. The bound is strict: the answer must be larger than it. Use whole-number entries; commas separate cycle lengths.
Two original transfers
Satellite exercise: orbital periods 6, 10, and 15 days synchronize after:
lcm(6,10,15) = 30 days
original Exercise 6 — intended nontrivial reading: a number leaves remainder 1 when divided by every integer from 2 through 9. If the answer is required to be greater than 9, then:
N − 1 is divisible by 2,3,4,5,6,7,8,9
lcm(2,3,4,5,6,7,8,9) = 2520
N = 2520 + 1 = 2521
As printed, the value 1 also satisfies the remainder condition. The worked example appears to intend the least nontrivial value greater than all the listed divisors; this lesson states that extra condition openly.
Transfer checkpoint
Need a hint? Start here
Subtract the common remainder before finding a multiple. Apply the strict lower bound.
Check why it works
The lower bound greater than 9 excludes the otherwise valid value 1. Satellite return 30 days; LCM 2520; least permitted number 2521.
Mission 7
Not completeUse the LCM inside consecutive-number sums
Worked example 6 asks for the least positive natural number that can be written as the sum of 9, 10, and 11 consecutive positive natural numbers.
9 consecutive numbers
There is one middle number. The sum is:
9 × middle
So the total is divisible by 9.
10 consecutive numbers
Pair the first with the last, the second with the next-to-last, and so on. There are:
10 ÷ 2 = 5 equal pairs
So the total is divisible by 5.
11 consecutive numbers
There is one middle number. The sum is:
11 × middle
So the total is divisible by 11.
original result and construction
lcm(5, 9, 11) = 495
Divisibility gives a candidate. Writing all three sequences proves that 495 really works.
original checkpoint
Need a hint? Start here
Pair the first and last terms. Divisibility gives a candidate; a sequence proves it works.
Check why it works
The total must be a multiple of 5,9,11. The first candidate is 495, but it still needs three constructions. Five pairs; total 495; ten-term sequence starts at 45. The nine-term sequence is 51–59 and the eleven-term sequence is 40–50; each totals 495.
Mission 8
Not completeTest consecutive-sum representations instead of assuming them
For a total T written as k consecutive positive integers beginning at a:
Where does the formula come from? The first and last terms add to a + (a+k−1) = 2a+k−1. Two copies of the sequence can be paired into k equal sums, so 2T = k(2a+k−1). Therefore a = (2T/k − k + 1)/2. A representation works only when a is a positive whole number.
T = k(2a + k − 1) ÷ 2
original Exercise 1
Four consecutive natural numbers have sum 54:
12 + 13 + 14 + 15 = 54
Their least common multiple is:
lcm(12,13,14,15) = 5460
original Exercise 2 — “lucky numbers”
A lucky number is the product of three consecutive positive integers whose middle number is a perfect square.
3×4×5 = 60
8×9×10 = 720
A middle value of 1 would include zero, so it is not allowed. The next square middle is 16, and 15×16×17=4080 exceeds 2007. Later square middles give still larger products. Therefore, the LCM of all lucky numbers below 2007 is:
lcm(60,720) = 720
Transfer checkpoint
Need a hint? Start here
Solve for the first term and check that it is a positive whole number.
Check why it works
12+13+14+15=54. Square middles 4 and 9 give 60 and 720; middle 16 already gives 4080. First number 12; LCM 5460; lucky-number LCM 720.
Mission 9
Not completeIndependent synchronization workshop
Correct all eight questions to complete the workshop. Use the laboratories when you need to check a pattern.
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
Count circle positions once and add calendar days after Day 0. Answers: 216; 6; 21.6; 72; May; 30; 495; 5460.
Mission 10
Not completeExit ticket
Complete all ten missions and answer all five exit items correctly to earn your certificate.
0 / 5
LCM
Cycle & Consecutive-Sum Navigator
This certifies that a determined mathematician can synchronize repeating distances and dates, count shared marks without double-counting, and verify consecutive-number representations.
Lesson 19.3 • 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
Count 90÷18 + 90÷30 −1 = 7 distinct positions. Answers: 90; 7; 30; 495; 5460.
Learning notes and instructional additions
The four-consecutive-number, lucky-number, shared-remainder, and satellite extensions follow Exercises 1, 2, 6, and 13 on pages 115–116.Exercise 6, as printed, also admits the trivial value 1; the lesson labels the greater-than-9 condition as an intended nontrivial reading rather than silently treating it as original text.The general spacing, calendar, shifted-cycle, and consecutive-sum laboratories are added instructional scaffolds derived from those original structures.