Follow Remainder Cycles
第16讲 余数问题 · 数列、循环小数与周期位置
Shrink enormous sequence terms and long decimal expansions into short repeating remainder states. Then jump directly to a distant term, digit, sum, or route position.
Learn it on your own
Work through one or two missions at a time. First predict an answer on paper, then use a laboratory to test it. If stuck, open the local hint and then the reasoning. Try the workshop yourself before choosing a worked review after an attempted check. Every explanation you need is on this page; external chat is optional.
Always check: whole-number quantities, positive divisor, and 0≤r<d. A zero quotient is possible when N<d; a zero remainder means exact division. If a question asks for positive numbers, exclude zero.
A two-term rule needs a complete state
When the next term depends on the previous two terms, one repeated remainder is not enough. The future is determined by the ordered pair of consecutive remainders.
original-style rule
The modulus is the divisor used for remainder calculations. For remainder work, reduce each result to a number from 0 through modulus−1. The symbol ≡ means “has the same remainder as.”
Same current value, different future
Under the worked example rule modulo 6:
The current remainder is 3 in both states, but the previous remainder changes what comes next.
Ordered-pair state machine
The order is (previous, current); the new pair is (current, next). If you enter numbers outside the remainder range, the tool first replaces them with their remainders.
Check the complete-state idea
Need a hint? Try this first
Remember all information needed for the next step.
Read the reasoning, then explain it yourself
When the next term uses two previous terms, keep the ordered pair. Under 2(a+b)+1 modulo 6, (1,3) gives 3 next, while (2,3) gives 5: one current remainder 3 is not enough.
Worked example 6: generate the sequence and its remainders
The sequence begins 1, 3. From the third term onward, each term is twice the sum of the previous two terms, plus 1. Find the remainder of the 2,014th term after division by 6.
| Term n | Exact value aₙ | Remainder mod 6 | Remainder pair ending here |
|---|
Read the generated sequence
Need a hint? Try this first
Generate small remainders alongside the terms.
Read the reasoning, then explain it yourself
The terms start 1,3,9,25,69. Their remainders mod 6 are 1,3,3,1,3. Use 2(previous+current)+1 at every step.
Prove the cycle by returning to the same ordered pair
The sequence of single remainders looks repetitive, but the proof comes from the state pairs. Once the same ordered pair returns under the same rule, every later state must repeat.
Next remainder: 3
Next remainder: 1
Next remainder: 3
Complete the proof
Need a hint? Try this first
Return to the entire starting pair.
Read the reasoning, then explain it yourself
(1,3)→(3,3)→(3,1)→(1,3). The same rule now repeats the same states, proving period 3.
Jump directly to the 2,014th term
Once the period is proved, the huge term itself is unnecessary. Divide the term number by the period and use the position inside the repeating block.
Distant-term jumper
Finish the worked solution
Need a hint? Try this first
Use a one-based cycle position.
Read the reasoning, then explain it yourself
2014=3×671+1, so choose position 1 in the block 1,3,3. If the index remainder is 0, use the last position, not position 0.
General recurrence-remainder laboratory
Track only remainders, record ordered pairs, detect the first repeated state, and jump to any target term. The laboratory also handles initial terms before the repeating cycle. These are called warm-up terms. “Equivalent term” means an earlier term with the same remainder, not necessarily the same exact value.
Sequence 4, 5, 9, 14, 23, …
Each term is the sum of the previous two. Modulo 3, the ordered-pair state repeats after eight terms:
Since:
the 2,014th term has remainder:
Use the transfer recurrence
Need a hint? Try this first
Reduce the starting terms before continuing.
Read the reasoning, then explain it yourself
Starting 4,5 and adding the previous two gives mod-3 cycle 1,2,0,2,2,1,0,1. The pair returns after eight steps. 2014=8×251+6, so use position 6: remainder 1.
original Guided Practice 6: sum the first 100 digits after the decimal point in 1 ÷ 44
Long division produces a short nonrepeating prefix and then a repeating block. Count the prefix once, then count complete copies of the cycle.
| Digit position | Remainder before ×10 | Value after ×10 | Next digit | New remainder |
|---|---|---|---|---|
| 1 | 1 | 10 | 0 | 10 |
| 2 | 10 | 100 | 2 | 12 |
| 3 | 12 | 120 | 2 | 32 |
| 4 | 32 | 320 | 7 | 12 |
Remainder 12 has returned, so the two-digit block 27 now repeats.
Count the structure
Add the digits
Complete the worked example calculation
Need a hint? Try this first
Separate the nonrepeating prefix.
Read the reasoning, then explain it yourself
Long division gives 1÷44=0.022727…: after the decimal point, prefix 02 precedes repeated block 27. The first 100 digits contain 02 and 49 blocks of 27. Their sum is 2+49×9=443.
General repeating-decimal digit and sum laboratory
Enter only the digits after the decimal point: first the nonrepeating prefix, then one repeating block. Leave the prefix empty if repetition starts immediately. Leading zeros count as real positions; position 1 is the first digit after the decimal point.
Use a new repeating block
Digit positions start at 1 immediately after the decimal point. Do not count the zero before the point.
For 0.36 = 0.363636…:
Need a hint? Try this first
Count complete blocks, then a partial block.
Read the reasoning, then explain it yourself
For 0.363636…, the repeating block is 36. Position 20 is the second digit, 6. The first 12 digits are six blocks, sum 6×9=54.
Align two periodic sets of route positions
Periodic positions can be treated like remainder cycles. A location must belong to both sets at the same time.
Outbound and return rest stops
A cyclist travels from A to B, 950 km apart. On the outward trip, the cyclist rests every 90 km. After resting at B, the cyclist returns along the same road and rests every 100 km. Count only stops strictly between A and B; the endpoints do not count as common rest points. Exactly one return rest point is also an outward rest point.
Outward positions from A
Return positions measured from A
Route-position alignment laboratory
All distances are in kilometres measured from A. Counts and common positions exclude A and B. The diagram shows at most the first 36 distinct stop positions from A; the numerical results count every stop.
Verify the worked example location
Need a hint? Try this first
Measure both journeys from the same endpoint.
Read the reasoning, then explain it yourself
A shared rest point x satisfies x=90a and 950−x=100b for positive whole a,b. Testing the outward stops gives x=450: five outward intervals and five return intervals.
A huge number made only of ones
A number contains exactly 2013 digits, all 1: 111…111. Find its remainder when divided by 13. Do not try to type the whole number.
Hint: append one digit to the remainder
Appending 1 changes a number N into 10N+1. If N leaves remainder r modulo 13, the new number leaves the same remainder as 10r+1. Start with remainder 0 before any digits are appended.
Prove the cycle and jump
| Number of ones | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Remainder mod 13 | 0 | 1 | 11 | 7 | 6 | 9 | 0 |
For example 10×11+1=111=13×8+7. After six digits we return to remainder 0. Here the next step depends on only one remainder, so this return proves a six-step cycle. Since 2013=6×335+3, select the three-digit remainder: 7.
Use Check Mission 8 to check this application and the route questions together.
Remainder-cycle workshop
Correct all eight problems to complete the workshop.
Workshop strategy reminder
Write the division equation or identify the repeated change first. Use the nearby mission hints for the matching method. For “all possible” answers, list the candidates systematically and explain why no other candidate can pass. Check every answer in the original story.
Objective exit ticket
Try these questions with the examples covered. Complete Missions 1–8, earn 8/8 in the workshop, and earn 5/5 here to unlock your certificate. Reflection is optional and ungraded.
Optional reflection — not automatically graded
Certificate of Completion
This certifies that
completed Lesson 16.6 — Follow Remainder Cycles in Sequences and Decimals
Worked explanation for the exit ticket
20=3×6+2 selects 4; 21=3×7 selects the last value 1. After prefix 15, ten positions hold five copies of 56. The 12th digit is 6, and the sum is 1+5+5×(5+6)=61. At x=90, the distances are 90=3×30 and 170−90=80=2×40; testing 30,60,90,120,150 shows no other match.
If you used this explanation, close it and solve the questions again on paper. Explain why each remainder is smaller than its divisor.
Optional extra: Find Full Time Timestamps. This bonus does not affect lesson completion.