Follow Recursive Digit and Decimal Cycles
第21讲 数字和 · Lesson 21.4
Generate only enough digits to expose the repeating state. Then replace a distant position or a very long digit sum with a short cycle calculation.
Track the complete state, not one repeated digit
When the next digit depends on the previous two digits, the ordered pair of current digits determines the future.
One digit is not enough
In a rule such as “multiply the previous two digits and keep the units digit,” the next digit depends on both entries.
A cycle is proved when the same ordered pair returns under the same rule.
One-step state machine
Checkpoint
Need a hint for Mission 1?
The ordered pair determines the next digit and the next pair.
Check your method
From(8,9),8×9=72 gives next digit2 and next state(9,2). Two digits;ordered pair;preperiod;2.
Predict before expanding a cycle
For a two-term rule, write the previous and current digit together. Returning to this ordered pair forces the same next digit, and then the same continuation. Before summing a long string, mark the opening, bracket one full cycle, and circle the leftover digits. For the block 230769, an extra zero can add a position without adding to the sum; this is why the segment problem can have two lengths.
Reveal the worked example product recurrence
Start with 1, 9, 8, 9. From the fifth digit onward, keep the units digit of the product of the previous two digits.
original rule
Sequence explorer
(2,8) returns after six digits, so every later digit follows the same six-step route.Rebuild the worked example cycle
Need a hint for Mission 2?
Use the six-digit block starting immediately after1989.
Check your method
The next digits are2,8,6,8,8,4, then the same pair(2,8) returns. Fifth digit2;block286884;length6;sum36.
Sum the first 2,014 original digits
Keep the four-digit opening separate, then divide the remaining positions into complete six-digit cycles.
original breakdown
Jump to any target length
Checkpoint
Need a hint for Mission 3?
The first2014 digits include the four-digit opening.
Check your method
Opening sum27;2014−4=2010=335×6. 27;2010;335;27+335×36=12087.
Complete the worked example sum-recurrence practice
Start with 9, 2, 1, 3. Each new digit is the units digit of the sum of the preceding two digits.
Practice rule
Cycle audit
Checkpoint
Need a hint for Mission 4?
Use the addition rule and the block starting at the first digit.
Check your method
921347189763 repeats with sum60. There is no preperiod.100=8×12+4. Block921347189763;period12;sum60;total495.
Use a general recursive-digit cycle laboratory
Enter the given beginning, choose the rule, and let the laboratory detect the preperiod, repeating block, distant digit, and first-N sum.
Recursive digit laboratory
21 transfer
The list begins with 3, then repeats the six-term block 2, 1, 3, 5, 7, 6.
Checkpoint
Need a hint for Mission 5?
Use the named fixed rules, even if you change the laboratory.
Check your method
Source50-digit sum:27+7×36+(2+8+6+8)=303. The separate list starts with3 once, then a block totaling24. 303;preperiod length0;24;3+33×24+2=797.
See how the decimal blocks for sevenths rotate
The six fractions 1/7 through 6/7 use rotations of the same six digits. Every complete cycle has the same digit sum.
The repeating family
Explore one fraction
Checkpoint
Need a hint for Mission 6?
Count only digits after the decimal point.
Check your method
Every block has six digits totaling27.4/7 starts571428;6/7 starts857142. 6;27;571428;8571.
Recover the numerator from a distant decimal digit sum
Subtract the contribution of every complete six-digit cycle. The leftover positions identify which rotation was used.
Worked example 4
Only the two-digit beginning 57 has digit sum 12, so the worked example fraction is 4/7.
Guided Practice 3
Only the four-digit beginning 8571 has digit sum 21, so the practice fraction is 6/7.
Inverse rotation solver
| a | Block | Leftover prefix | Prefix sum | Predicted total | Status |
|---|
Checkpoint
Need a hint for Mission 7?
The unknown numerator a is an integer1–6 in the fraction a/7.
Check your method
For2012 places,335 full cycles contribute9045;9057−9045=12. For2008 places,334 cycles leave21 to come from four digits. 335;12;a=4;practice a=6.
Test two original extensions carefully
A target digit sum may allow more than one segment length, while a recursive number sequence may be recovered from a later term.
Contiguous segment of 3/13
3/13=0.230769…. Search for consecutive digits whose sum is 2003. A segment may start at any decimal position, may cross block boundaries, and includes no gaps. Count digits after the decimal point; the initial 0 is not part of a segment.
Fibonacci-type transfer
The first two terms are positive whole numbers x and y with x<y. Each later term is the sum of the preceding two. In the fixed source problem the fifth term is 21. Find the eighth term.
Checkpoint
Need a hint for Mission 8?
The fixed segment uses230769 and sum2003; the fixed number recurrence has fifth term21.
Check your method
74 full blocks sum1998. The remaining sum5 comes from23, or230 including the zero, giving446 or447 digits. For the recurrence2x+3y=21 with positive integers x<y forces x=3,y=5. Cycle sum27;446;447;eighth term89 in3,5,8,13,21,34,55,89.
Try a new problem before the workshop
The repeating block is 230769. What is the sum of the first 14 digits?
Compare your reasoning after trying
Two full blocks contribute 2×27=54; the next two digits contribute 2+3=5. Total: 59.
Independent workshop
Correct all eight answers to complete the workshop.
Need a hint for Mission 9?
Keep each sequence’s opening and rule separate.
Check your method
Sum opening + complete blocks + leftover digits; count a zero as a position even though it adds nothing. 2;6;12087;12;495;4;797;446.
Objective exit ticket
Answer all five correctly and complete the other nine missions to unlock the certificate.
Optional reflection — not automatically graded
Recursive Digit & Decimal Cycle Navigator
This certifies that Student completed all ten missions and can detect full recurrence states, use preperiods and cycles, and recover information from repeating decimals.
Chapter 21 • Digit Sums
Need a hint for Mission 10?
For the final item,11=2×4+3.
Check your method
Two1234 blocks contribute20; the final123 contributes6. 36;303;6;797;26.