Replace large numbers with small remainder labels, make one case for every possible remainder, and turn each case into an explicit construction. You will audit stamp purchases, subset sums, package sizes, adjacent sums, and parity pairs.
When a positive integer N is divided by m, it belongs to exactly one of the complete cases:
For sums, replace each number by its remainder:
“43 mod 5 = 3” means 43 leaves remainder 3 when divided by 5: 43 = 5 × 8 + 3. “a ≡ b (mod 5)” means a and b have the same remainder. Only 0, 1, 2, 3, 4 are possible remainders, so these classes cover every integer exactly once.
If two numbers leave the same remainder upon division by m, their difference is divisible by m.
The possible remainders modulo m are 0 through m−1, never m.
Package counts are whole numbers, including zero. A purchase using only one size is allowed. For each count of the larger package, subtract its contribution and test whether the remainder is a nonnegative multiple of the smaller size.
If x is the number of 3-jiao stamps, then:
So the amount’s remainder modulo 5 determines the least number of 3-jiao stamps compatible with maximizing the 5-jiao stamps.
| Final digits | Remainder mod 5 | 3-jiao stamps per student | Students | Contribution |
|---|
Fix the larger package count; the remaining amount must be a nonnegative multiple of the smaller size.
| Amount | Remainder | Smaller items | Preferred items | Check |
|---|
Try each remainder class, then use a repeatable construction to cover all larger values in that class.
From 1995, 2000, 2001, 2003, 2007, choose a nonempty subset whose sum leaves remainder 7 upon division by 9. Each value is used at most once, and order does not matter.
Only the small remainder labels matter when checking the final remainder.
Add the remainders of the selected cards, then take the remainder again.
Group by remainder before counting subsets. A subset has one exact selection, even if several selections have the same sum.
Use only 3-kg packages.
Use two 5-kg packages, then fill the rest with 3-kg packages.
Use one 5-kg package, then fill the rest with 3-kg packages.
Find the first reachable amount in each remainder class. Adding a whole package preserves reachability.
Find four consecutive reachable totals for sizes 4 and 7. Adding another 4-kg package then reaches every later total.
C(41,2), also written “41 choose 2,” counts pairs of different items. There are 41 choices for the first and 40 for the second. Each pair is listed in two orders, so divide 41 × 40 by 2. In the parity problem, count odd–odd and even–even pairs separately, then add.
Three cards have positive whole-number values at most 10. Each round, three people each receive one card. After the same number of rounds their totals are 13, 15, and 23. Find the card values and number of rounds.
Every round uses all three cards. Add the three final totals, then consider which whole-number round counts could divide that total. Test the bounds on card values.
Write the cards as a ≤ b ≤ c ≤ 10. List a from 1 to 5; for each a, list b from a to floor((17−a)/2), keeping c=17−a−b ≤ 10. Test three-card sums for each person's total. Only 3, 5, 9 works.
| Round | Person A | Person B | Person C |
|---|---|---|---|
| 1 | 3 | 9 | 5 |
| 2 | 5 | 3 | 9 |
| 3 | 5 | 3 | 9 |
| Total | 13 | 15 | 23 |
Each row uses each card once, so the construction really is possible.
Sequence:
A prefix sum adds terms from the start: P₀=0, P₁=1, P₂=1+4, and so on. A nonempty consecutive group from term i+1 through term j sums to Pⱼ−Pᵢ. It is divisible by 11 exactly when those two prefix sums have the same remainder modulo 11.
An even sum uses two numbers of the same parity:
An even sum comes from two odd numbers or two even numbers. Divide ordered pair counts by two.
Correct all eight questions to complete the workshop.
Answers: 7; 84; 4; 4; 30; 18; 7; 1600. For 47 jiao use four 3-jiao stamps and seven 5-jiao stamps. Package totals 18–21 are all possible. For pairs, add 41×40÷2 and 40×39÷2.
Earn 5 out of 5. The certificate also requires Missions 1–9 to be complete.
Answers: 5; 1; 30; 8; 90. Use 68=9×7+5 and 33=3+6×5. Triples: 3 same-class plus 27 mixed-class choices. Package totals 8,9,10 begin the three remainder classes. For 1–20, ten odds and ten evens give 2×(10×9÷2)=90 unordered pairs.
This certifies that the learner can build complete remainder cases, count residue patterns, and construct a valid result in every case.