Find What Stays the Same
第16讲 余数问题 · 不变量与余下的一项
Use a remainder as a fingerprint: repeated operations may change the total without changing its remainder, and grouped totals can reveal the one item left behind.
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 remainder can be a fingerprint
If every allowed operation changes a total by a multiple of the same number, the remainder under division by that number cannot change.
Read the original example: start 6, add 5 per action, take 4 actions
Need a hint? Try this first
Calculate the change, not the whole total.
Read the reasoning, then explain it yourself
Adding 5 changes the quotient on division by 5 but preserves the remainder. For example 6,11,16 all leave 1.
Worked example 4: repeated paper cutting
A sheet is first cut into 6 pieces. Later, any selected pieces are each cut into 6 smaller pieces and returned. Which of 2012, 2013, 2014, 2015, and 2016 can be the total after an operation?
One selected piece
Why the remainder stays fixed
The first total is:
Every later change is a multiple of 5, so every reachable total also leaves remainder 1 when divided by 5.
Complete the worked example reasoning
Need a hint? Try this first
One piece disappears and six appear.
Read the reasoning, then explain it yourself
Each selected piece adds 6−1=5 pieces. Totals have form 6+5k. Among 2012–2016 only 2016 leaves 1 mod 5; selecting one piece each time constructs it in 402 further cuts.
A matching remainder is necessary; a construction proves possible
A remainder test can reject impossible totals. To prove a matching total is truly reachable, show a legal sequence of operations.
Five sheets, each chosen sheet becomes seven
The table starts with 5 stiff-paper sheets. In any round, choose one or more sheets, cut each chosen sheet into 7 smaller sheets, and return them. Can the table contain exactly 2015 pieces after a round?
Necessary condition
A candidate must leave the same remainder as the starting total:
Construction
A concrete legal plan reaches it:
Separate the two jobs
Need a hint? Try this first
Subtract the starting total.
Read the reasoning, then explain it yourself
Replacing one sheet by seven adds 6. Starting with 5, the increase needed is 2015−5=2010. Choose one sheet in each of 335 rounds: 5+335×6=2015. This legal plan proves possibility.
General operation-invariant laboratory
When one old item is replaced by k new items, the total increases by k−1. This laboratory checks both the remainder condition and a one-item-at-a-time construction.
Use the laboratory result
Need a hint? Try this first
Find the net increase from replacing one item.
Read the reasoning, then explain it yourself
Start with 8 items. Replacing one by four adds 3, so reaching 50 takes (50−8)÷3=14 actions. Reaching 49 would require 41÷3 actions, which is not a whole number. If no items exist at the start, a replacement cannot begin.
Worked example 5: identify the leftover box
Six boxes weigh 15, 16, 18, 19, 20, and 31 kg. Two customers buy five boxes. One customer’s total is twice the other’s. Find the weight left in the shop.
Turn the ratio into a divisor
The five purchased boxes weigh:
So their total is divisible by 3.
Use the whole total
The purchased total has remainder 0, so the leftover must carry the whole total’s remainder 2.
Complete the worked example proof
Need a hint? Try this first
Filter first, build second.
Read the reasoning, then explain it yourself
The boxes total 119, leaving remainder 2 mod 3. Only box 20 matches. Without it, 15+18=33 and 16+19+31=66, which proves the required ratio.
original Guided Practice 5: use group sizes and a ratio
Cards show 1193, 1258, 1842, 1866, 1912, and 2494. A takes 3 cards, B takes 2 cards, and C takes 1 card. A’s card total is twice B’s. Find C’s card.
Remainder filter
A and B together have 3 ratio parts, so their combined total is divisible by 3. C’s one card must leave remainder 2.
Verify the actual grouping
=6248
=3124
The unused card is 1193.
Complete the card reasoning
Need a hint? Try this first
Use both the ratio and the card counts.
Read the reasoning, then explain it yourself
The total is 10565, remainder 2 mod 3, so the unused card is 1193. Three cards 1842+1912+2494 total 6248; two cards 1258+1866 total 3124. This verifies the 2:1 ratio and the required counts.
Leftover finder: filter first, then verify a partition
The remainder condition identifies candidates. A complete search then checks whether the remaining items can actually be split into the stated group sizes and ratio.
Candy bags
Six bags contain 3, 4, 5, 7, 9, and 13 candies. Xiaohua takes 2 bags. Xiaoming takes 3 bags containing twice as many candies in total. One bag remains.
The unique verified leftover is 5: Xiaohua can take 3+9=12, while Xiaoming takes 4+7+13=24.
Complete the transfer
Need a hint? Try this first
A matching remainder is only a candidate.
Read the reasoning, then explain it yourself
For bags 3,4,5,7,9,13, the whole total is 41. Removing 5 leaves 36: 3+9=12 and 4+7+13=24. For any new list, the laboratory also checks an actual partition.
Enter 3–12 positive whole-number item values. Every entry is one separate item, even when two entries have the same value. Leave exactly one item; use every other item exactly once in two nonempty groups. The ratio compares the groups’ totals, not their numbers of items. A count of 0 lets the laboratory choose that group’s size. Ratios are reduced first: 2:4 means 1:2, so the filter uses modulus 3, not 6. A partition must still satisfy the original ratio.
Two more invariants hidden in the chapter exercises
A ratio may create a five-part total, and repeating a digit block creates a fixed quotient independent of the chosen block.
Three buyers and one leftover box
Eight candy boxes contain:
The one box taken by A must leave remainder 1 modulo 5, so it is 31. A grouping that proves this works is B: 30+44=74, C: 17+24+33=74, and D: 9+28=37. Each box is used once; A takes the remaining box containing 31 candies.
Repeat a two-digit block four times
Because the quotient is always 1010101, its remainder after division by 9 is always:
Finish both invariants
Need a hint? Try this first
Count ratio parts or repeated place values.
Read the reasoning, then explain it yourself
The eight boxes total 216. Totals in ratio 2:2:1 use five parts, so leftover 31 matches remainder 1 mod 5; 30+44=74, 17+24+33=74, 9+28=37 verify it. Repeating a two-digit number x four times gives x×1010101; dividing by x leaves 1010101, whose digit sum 4 gives remainder 4 mod 9.
Apply remainders to a long sum
Find the remainder when the sum of the first 2014 positive multiples of 6 is divided by 9. The sum starts 6+12+18+24+… . Predict a short method before opening the hint.
Hint: group three consecutive terms
The remainders modulo 9 repeat 6,3,0. Each complete block sums to 9 and contributes remainder 0. Why does it repeat? Three steps add 18, a multiple of 9.
Worked reasoning
2014=3×671+1. There are 671 complete three-term blocks and one remaining term. That term has remainder 6, so the whole sum leaves 6. This avoids calculating the large total.
Use a remainder table for a tournament
Four players have labels 101,126,173,193. Each pair plays as many games as the remainder of the sum of their labels divided by 3. A remainder of 0 means they play no games. What is the greatest total number of games played by any one player?
Hint: reduce each player’s label first
The labels have remainders 2,0,2,1. Add the two remainders for a pair and reduce again modulo 3.
Worked table
| Player | 101 | 126 | 173 | 193 | Total |
|---|---|---|---|---|---|
| 101 | — | 2 | 1 | 0 | 3 |
| 126 | 2 | — | 2 | 1 | 5 |
| 173 | 1 | 2 | — | 0 | 3 |
| 193 | 0 | 1 | 0 | — | 1 |
Player 126 plays the most: 5 games. Each row counts one player’s games. Adding all row totals would count every game twice.
Use Check Mission 8 to check these two applications along with the earlier questions.
Invariants and leftovers workshop
Use the change per action, the fixed remainder, ratio-part totals, and a final construction or partition check. Correct all eight answers to complete this 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.
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 16.5 — Find What Stays the Same: Invariants and Leftovers
Worked explanation for the exit ticket
Replacing one piece by 7 adds 6. (100−4)÷6=16; 101−4 is not divisible by 6. The items total 34, remainder 1 mod 3. Candidates 4,7,10 match. Leaving 4 would need a pair totaling 10, impossible; leaving 10 would need a pair totaling 8, impossible. Leaving 7 gives pairs 4+5=9 and 8+10=18. The smaller total is 9.
If you used this explanation, close it and solve the questions again on paper. Explain why each remainder is smaller than its divisor.