Transform and Align Remainders
第16讲 余数问题 · 余数的变换与对齐
Replace large numbers with small remainders, combine them safely, rebuild nested division stories, and line up several conditions with one useful adjustment.
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.
Replace a number by its remainder
For one fixed divisor, numbers with the same remainder sit in the same remainder class. Their difference is a whole multiple of the divisor.
Compact notation, introduced only after the words are clear: 38 ≡ 3 (mod 7).
Read the remainder story
Need a hint? Try this first
Remove full turns of the wheel.
Read the reasoning, then explain it yourself
38=7×5+3, so 38 and 3 leave the same remainder on division by 7. The symbol “mod 7” means “remainder when divided by 7.”
Add and subtract remainders, then reduce again
The raw sum or difference of two remainders may fall outside the allowed range. Reduce it to a number from 0 through d − 1.
Two-number remainder laboratory
For subtraction, a negative intermediate result means moving backward on the remainder wheel. Add the divisor until you reach 0 through d−1. This convention also handles a negative difference: −9 = 8×(−2)+7, so its remainder modulo 8 is 7.
Add
24 + 18 = 42, and 42 ÷ 7 leaves 0.
Subtract
24 − 18 = 6, and 6 ÷ 7 leaves 6.
Practice Question 3
If a leaves remainder 3 and b leaves remainder 4 when divided by 7, then:
Complete the transformation
Need a hint? Try this first
Reduce again after the operation.
Read the reasoning, then explain it yourself
3+4=7 gives remainder 0 modulo 7. For subtraction, 3−4=−1; add 7 to obtain the valid remainder 6.
Multiply a remainder and reduce again
If N leaves remainder r, then kN has the same remainder as k×r. The product still needs to be reduced.
Practice Question 4
A number leaves remainder 3 when divided by 7. What remainder does three times that number leave?
Multiplier laboratory
Complete the worked example test
Need a hint? Try this first
Multiply the small remainder.
Read the reasoning, then explain it yourself
If N=7q+3, then 3N=21q+9=7(3q+1)+2. The final remainder is 2. Do not assume division of remainders works this way: 8 and 3 both leave 3 mod 5, but halving 8 gives 4, not 1.5.
Worked example 3: rebuild a nested quotient
A remainder condition on the quotient is not a remainder condition on the original number. Write two division equations and substitute.
Complete practice problem
A nonnegative whole number N leaves remainder 2 when divided by 3. The quotient from that division leaves remainder 3 when divided by 4. What remainder does N leave when divided by 6?
Generate examples without changing the final remainder
Complete the worked example derivation
Need a hint? Try this first
Name the inner quotient first.
Read the reasoning, then explain it yourself
q=4t+3. Substitute into N=3q+2: N=12t+11=6(2t+1)+5. The final remainder is 5.
Guided Practice 3: align one expression with another
Rewrite the larger expression as a whole multiple of the divisor plus a small leftover.
Complete original practice
When b is divided by a, the quotient is 3 and the remainder is 3. When c is divided by a, the quotient is 9 and the remainder is 11. Therefore a>11. The division equations are:
What remainder does c leave when divided by b?
Scale b
Compare with c
Check many values of a
Complete the alignment
Need a hint? Try this first
Compare c with three copies of b.
Read the reasoning, then explain it yourself
The original divisions require a>11. Since 3b=9a+9, c=3b+2. Also b≥39, so 0≤2<b and the remainder is 2.
Align several conditions with one shared offset
Sometimes every remainder sits the same distance below its divisor. Add that distance once, and all conditions become exact divisibility conditions.
Practice Question 6
Find the smallest positive whole number N that leaves remainders 1, 2, 3, and 4 when divided by 3, 4, 5, and 6 respectively.
| Divisor d | Required remainder r | Gap d−r |
|---|---|---|
| 3 | 1 | 2 |
| 4 | 2 | 2 |
| 5 | 3 | 2 |
| 6 | 4 | 2 |
Test a candidate
Complete the shared-offset reasoning
Need a hint? Try this first
Look at each gap divisor−remainder.
Read the reasoning, then explain it yourself
All four gaps are 2, so N+2 is divisible by 3,4,5,6. Their least common multiple is 60. The least positive solution is 58. Check all four divisions.
Use a simultaneous-remainder search laboratory
A shared offset is elegant when it exists. A complete search handles both aligned and unaligned systems—and can prove impossibility.
Enter matching lists of divisors and remainders
Match entries in order: the first remainder belongs to the first divisor, and so on. Divisors must be at least 2. This tool finds the smallest nonnegative solution, including zero when it works. If the question asks for a positive solution and the tool returns zero, use the LCM as the first positive solution. Periods above 200,000 are not searched.
Read three systems
Need a hint? Try this first
A common gap helps, but is not required.
Read the reasoning, then explain it yourself
For N=1 mod 4 and N=3 mod 5, test 1,5,9,13: 13 works. For divisors 5,7,9 and gap 4, N=315−4=311. A number cannot be odd (1 mod 4) and even (2 mod 6) simultaneously.
Work backward through repeated “divide into thirds” steps
This practice problem nests one remainder process inside another. Track the amount carried forward after each division.
original Exercise 12
A basket of apples is divided into three equal shares, with 2 apples left over. Every share at every stage contains at least one whole apple. Two full shares are taken and divided into three equal shares, again leaving 2. Two full shares from that division are taken and divided into thirds once more, again leaving 2. What is the least possible original number of apples?
Candidate simulator
First solutions found by complete search
Forward equations
Least successful chain
Complete the least chain
Need a hint? Try this first
Keep only two shares after each split.
Read the reasoning, then explain it yourself
Write N=3a+2, then 2a=3b+2, then 2b=3c+2. For positive shares, c must be even; the smallest is 2, giving b=4, a=7 and N=23. Forward check: 23→14→8, each leaving 2.
Remainder-transformation workshop
Correct all eight responses 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.
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
Remainder Transformation & Alignment Navigator
has completed Lesson 16.4 and can transform remainders under operations, rebuild nested division conditions, align several remainder clues, and audit repeated remainder processes.
Worked explanation for the exit ticket
4+5=9 leaves 3 mod 6. 4−5=−1; add 6 to get 5. 5×4=20 leaves 2 mod 9. Add 1 to align divisors 4 and 6: their LCM is 12, giving N=11. Substitute q: N=15t+7.
If you used this explanation, close it and solve the questions again on paper. Explain why each remainder is smaller than its divisor.
Further practice
The nested-quotient investigation and the b–c guided practice follow Chapter 16, Example 3 and Guided Practice 3 on. The repeated-thirds apple problem follows Chapter 16 Exercise 12 on page 100. The addition, multiplication, and shared-offset investigations follow Test 16 Questions 3, 4, and 6 on Practice .