Last Digits and Recursive Remainder Cycles
末位数字与递推余数周期
Huge powers and fast-growing sequences can be surprisingly small problems. Track only the remainder the question asks for—and, in a two-term rule, remember that the complete state is an ordered pair.
Keep only the digit the question needs
The units digit of a product depends only on the units digits of its factors. All earlier digits change multiples of 10, so they cannot change the final digit.
Last-digit product machine
8 × 7 = 56, so 38 × 47 ends in 6.
Why this works
Write the numbers as:
Their product is:
Every term except rs is a multiple of 10. Therefore only the last digits r and s control the last digit of the product.
Check the idea
Use the numbers written in each question, even if you changed the machine.
Build a power cycle
Repeated multiplication creates a repeating pattern of units digits. The exponent tells which position in the cycle to use.
| Base's last digit | Repeating units digits of positive powers | Period |
|---|
Worked example: find the units digit of 72015
The worked example asks for the last digit of a product containing 2,015 factors of 7. We never calculate the full power.
Step 1 · Use the cycle
Step 2 · Divide the exponent by the period
Step 3 · Read the third cycle item
Extra practice: find the units digit of 23425
The worked example defines 23425 as 25 factors of 234. Only the base's units digit, 4, matters.
Replace the base by its last digit
The cycle has period 2.
Locate exponent 25
Remainder 1 means the first cycle item:
25 = 2 × 12 + 1 → first cycle item → 4
Use a general last-digit laboratory
Choose a whole-number base and a positive whole-number exponent, and whether to track the last one or two digits. “Mod 10” means keep the remainder after division by 10; “mod 100” means keep the remainder after division by 100. For two digits, write a leading zero when needed, such as 07. The process repeats because there are only finitely many possible remainders.
For a two-term rule, the state is an ordered pair
If the next term depends on the previous two terms, one repeated digit is not enough. The same ordered pair returning guarantees that the same rule will produce the same future. “Ordered” means previous term first, current term second.
Pair-state machine
2 × 8 = 16 → next digit 6
Same current digit, different future
Under the product rule:
Both states end in 8, but they do not have the same future. A repeated ordered pair is the evidence needed for a cycle.
original exercise: a product-recurrence cycle
Start with 1, 9, 8, 9. Beginning with the fifth digit, each new digit is the units digit of the product of the previous two digits.
Generate enough terms to find a repeated pair
The pair (2, 8) begins at Terms 5–6 and returns at Terms 11–12.
2012 − 4 = 2008 = 334 × 6 + 4 → 4th cycle digit = 8
Sum recurrences and Fibonacci remainder cycles
The same ordered-pair method works when the next digit is a sum, and when we track Fibonacci numbers only by their remainders.
original exercise: 9, 2, 1, 3, …
From the third digit onward, use the units digit of the sum of the previous two digits.
100 = 8 × 12 + 4 → 100th digit = 3
Sum of the first 100 digits: 8 × 60 + (9 + 2 + 1 + 3) = 495. Each complete 12-digit block sums to 60.
Fibonacci remainders after division by 3
Start with 1, 1. Each new number is the sum of the previous two, but record only the remainder after division by 3.
The ordered pair (1, 1) returns after 8 terms, so the remainder cycle has period 8.
2015 = 8 × 251 + 7 → 7th cycle remainder = 1
Recursive-remainder laboratory
Enter whole-number starting values from 0 to 99. The lab first replaces each by its remainder for your chosen divisor (2–100). Term 1 is the first starting value’s remainder and Term 2 is the second. The fixed questions below use the two examples above, regardless of the lab settings.
original Exercise 5: growing sums, repeating parity
Row n contains 1,2,…,n. For rows 1 through 2,015, how many row sums are even? The first sums are 1,3,6,10,15,21,28,36: odd, odd, even, even, then repeat.
Why does the parity repeat?
The row sum is n(n+1)/2. It is even when n or n+1 is divisible by 4: rows n ≡ 0 or 3 modulo 4. Here “modulo” means “classified by the remainder.”
Count the even sums
2,015 = 503 × 4 + 3. Each full block has two even sums, and the first three rows of the next block add one. Answer: 503 × 2 + 1 = 1,007.
original Exercises 7–8: use endings to calculate and prove
Find the units digit of 12345678922. Then prove that 32000 + 42001 is divisible by 5.
Hint: track only the endings
Powers of 9 alternate 9,1; powers of 3 cycle 3,9,7,1; powers of 4 alternate 4,6. A whole number is divisible by 5 if its units digit is 0 or 5.
Worked solution
An even positive power of 9 ends in 1, answering the first question. Since 2000 is divisible by 4, 32000 ends in 1. The odd power 42001 ends in 4. Their sum ends in 5 and is therefore divisible by 5.
original Exercise 12: the best starting point
Start with 1,3. Every later term is the units digit of the product of the previous two. The sequence begins 1,3,3,9,7,3,1,3,… . Choose 2,017 consecutive terms to make their product as large as possible. What is the units digit of that largest product?
Hint: full blocks versus the extra term
The pair (1,3) returns after six terms, giving block 1,3,3,9,7,3. Divide 2,017 by six. The full six-term blocks have the same product no matter where they start.
Worked solution
2,017 = 336 × 6 + 1. Every full block has product 1 × 3 × 3 × 9 × 7 × 3 = 1,701. Only the extra term changes the product, so choose a starting point whose extra term is 9. The maximum is 1701336 × 9, whose units digit is 9.
Independent workshop
Try each new problem before opening a hint. Correct all 8 answers to complete this mission.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 3, 9, 7, 1. Exponent 41 uses cycle position 1 of 4, so the ending is 3.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 8, 4, 2, 6. Exponent 46 uses cycle position 2 of 4, so the ending is 4.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 9, 1. Exponent 37 uses cycle position 1 of 2, so the ending is 9.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 4, 6. Exponent 18 uses cycle position 2 of 2, so the ending is 6.
Hint
Multiply the pair and keep the units digit.
Worked explanation
7 × 8 = 56, so the next digit is 6.
Hint
Generate the remainders modulo three, tracking consecutive pairs until the starting pair returns.
Worked explanation
18 = 2 × 8 + 2. The second remainder is 1.
Hint
Keep only remainders after division by 100. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 07, 49, 43, 01. Exponent 6 uses cycle position 2 of 4, so the ending is 49.
Hint
Compute the sum, or locate 15 in the four-row parity pattern.
Worked explanation
15 × 16 ÷ 2 = 120, which is even.
Transfer exit ticket
Try each new problem before opening a hint. Correct all 5 answers to complete this mission.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 7, 9, 3, 1. Exponent 2030 uses cycle position 2 of 4, so the ending is 9.
Hint
Keep only remainders after division by 10. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 4, 6. Exponent 19 uses cycle position 1 of 2, so the ending is 4.
Hint
Add the remainders, then divide by five and keep the remainder.
Worked explanation
2 + 4 = 6, which leaves remainder 1 modulo 5.
Hint
Generate the Fibonacci remainders modulo three and look for the return of the complete starting pair.
Worked explanation
22 = 2 × 8 + 6. The sixth remainder is 2.
Hint
Keep only remainders after division by 100. Track one multiplication at a time until a remainder repeats.
Worked explanation
The positive-power endings repeat 09, 81, 29, 61, 49, 41, 69, 21, 89, 01. Exponent 12 uses cycle position 2 of 10, so the ending is 81.
Certificate of completion
Last-Digit & Remainder-Cycle Navigator
This certifies that the learner can find power-ending cycles, track complete pair states, and jump to distant terms without calculating enormous numbers.