Chapter 22 · Lesson 4

Turn Divisibility Claims into Remainder Classes

把整除断言转化为余数类

Use implications instead of testing every claim separately. Then turn “every pair has a divisible sum” into a small remainder-class problem.

Force the claims → build the number → compare the residue classes.

Worked example 4Guided Practice 310 interactive missionsWorks offline

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Read a worked example, predict what the lab will show, then test your prediction. Finish a number entry and press Tab (or choose another control) to update a lab. The local hints and explanations work without opening another website.

Useful words and symbols

Consecutive means following in order. Consecutive integers have gap 1; consecutive even or odd numbers have gap 2. Positive integers start at 1. A lab explicitly allowing zero includes 0 as well.

Remainder is what is left after making equal groups. “17 mod 5 = 2” and “17 ≡ 2 (mod 5)” both say that 17 leaves remainder 2 when divided by 5. A residue class groups numbers with the same remainder.

Coprime numbers have no common factor greater than 1. The LCM is the least positive common multiple. A self-inverse remainder pairs with itself to make a multiple: for example 5+5 is a multiple of 10. Lexicographic order compares the first differing digit, as dictionary order compares letters.

To prove an answer is the best possible, explain why a better one cannot work, then give an example that reaches your bound.

Mission 1

Learn which way a divisibility claim travels

A true divisibility claim travels down to divisors. A false claim travels up to multiples.

Not complete

True claim: travel down

N is divisible by 1212 is the larger requirement

→

N is divisible by 2, 3, 4, and 6Every divisor of 12 also works

False claim: travel up

N is not divisible by 55 fails

→

N is not divisible by 10, 15, 20, …No multiple of 5 can work

Implication explorer

Chosen labelShow labels through (at least the chosen label)

If the claim is true

If the claim is false

Checkpoint

Divisible by 12 guarantees divisible by 4? Not divisible by 5 guarantees not divisible by 20? Not divisible by 6 guarantees not divisible by 3? Divisible by 3 and 4 guarantees divisible by 12?

Mission 2

Test every adjacent pair in the practice problem

Students 2 through 15 each claim that one five-digit number is divisible by their own label. Exactly two claims are false, their labels differ by 1, and every other claim is true. A label d means the claim “N is divisible by d.”

Not complete

original EXAMPLE 4

Which adjacent pair can be wrong?

Choose a proposed false pair. The runtime checker asks whether a five-digit number can be divisible by every other label from 2 through 15 while failing both chosen labels.

First false label

Checkpoint

Number of feasible adjacent pairsFirst false labelSecond false labelFive-digit witness

Mission 3

Prove that labels 2 through 7 must be true

Assume one small label is false. Its double must also be false—but the two labels are not adjacent.

Not complete

Double-witness explorer

Suppose this label is falseSelected label: 4

original proof chain

If label n is false, every multiple of n is false.

For 2 ≤ n ≤ 7, label 2n also lies from 4 through 14.

The labels n and 2n differ by n, so they are never adjacent.

That contradicts “exactly two adjacent labels are wrong.”

Checkpoint

If label4 is false, what is its double, which must also be false?Are labels 4 and 8 adjacent?Smallest forced-true labelLargest forced-true label

Mission 4

Propagate the true claims and isolate the wrong pair

Once 2 through 7 are true, several larger labels become true automatically.

Not complete

Forced by least common multiples

Divisible by 2 and 5 ⇒ divisible by 10.

Divisible by 3 and 4 ⇒ divisible by 12.

Divisible by 2 and 7 ⇒ divisible by 14.

Isolated labels must also be true

Label 11 lies between true labels 10 and 12. Label 13 lies between 12 and 14. Label 15 could only form an adjacent false pair with 14, which is true.

So 11, 13, and 15 are true. Only the adjacent labels 8 and 9 remain.

Checkpoint

LCM of 2 and 5LCM of 3 and 4LCM of 2 and 7List both wrong labels (commas or spaces, either order)

Mission 5

Build the five-digit number from the true labels

The number must be a common multiple of every true label, so start with their least common multiple.

Not complete

original EXAMPLE 4

Prime-factor ledger

The true labels require the prime powers in 12, 5, 7, 11, and 13.

2^e 3^e 5^e 7^e 11^e 13^e

Checkpoint

Least common multipleRemainder when divided by 8Remainder when divided by 9Twice the LCM of the true labels

Mission 6

Audit any number against a line of divisibility claims

This added laboratory checks every label, groups consecutive false claims, and identifies the longest false run.

Not complete

Number to auditFirst labelLast label

Checkpoint for 60,060 and labels 2–15

Number of true claimsNumber of false claimsNumber of labels in the longest consecutive false runFalse labels

Mission 7

Turn pairwise divisible sums into residue classes

Choose distinct whole numbers, using each at most once. Suppose at least three selected numbers have the property that the sum of every two distinct selected numbers is divisible by m.

Not complete

Three-residue proof

a + b ≡ 0 (mod m)
a + c ≡ 0 (mod m)

Subtract the two congruences:

b ≡ c (mod m)

Now use the pair b and c. Since b and c have the same remainder, their sum has the remainder of2b:

2b ≡ 0 (mod m)

Self-inverse residue classes

A residue class can hold several selected numbers only when adding the residue to itself gives 0.

Modulus m

Also a≡−b≡b, since2b≡0; thus all three have the same remainder. Any additional chosen value must share it by pairing with a. Two-number exception: with exactly two selected values, complementary residues r and m−r may work. The “all values lie in one self-inverse class” conclusion needs at least three selected values.

Checkpoint

From a+b≡0 and a+c≡0, b is congruent to…Then 2b is congruent to…List all self-inverse residues modulo26 (commas or spaces, either order)List all self-inverse residues modulo15

Mission 8

Maximize the worked example selection by comparing residue classes

Guided Practice 3 asks for the largest subset of 1 through 1994 in which every pair has a sum divisible by 26.

Not complete

GUIDED PRACTICE 3

Only residues 0 and 13 can support three or more values

The two classes cannot be mixed, because 0+13 is not divisible by 26. Count each class and choose the larger one.

Range beginsRange endsDivisor m

One maximum selection

Checkpoint for the worked example range

How many numbers in1–1994 leave remainder0 on division by26?How many numbers in1–1994 leave remainder13 on division by26?Maximum selection sizeLargest selected value in the winning class

Try it on your own — a new problem

This extra practice has its own saved checkpoint. Try it, check your answer, then use a hint or the separate worked review.

Choose numbers from 1–30 so every pair of distinct chosen numbers has a sum divisible by 10. What is the largest possible number of choices?

Your answer

Explain your reasoning: Why must you check every pair in your proposed set, including pairs from different remainder classes?

Mission 9

Independent workshop

Correct all eight answers to complete the workshop.

Not complete

1. If N is divisible by 18, it is certainly divisible by:

2. If N is not divisible by 7, can it be divisible by 21?

3. For a five-digit positive integer N, exactly two labels d from 2 through 15 make false claims that d divides N, and those two labels differ by 1. List the false labels (either order):

4. Find N in the claim puzzle of question 3:

5. 60,060 leaves what remainder upon division by 8?

6. For three or more numbers whose pair sums are divisible by 26, list the two possible shared residue classes. One selected set uses just one class:

7. Multiples of 26 from 1 through 1994:

8. Maximum number of distinct choices from1–1994 whose every pair sum is divisible by26:

Workshop score0 / 8

Mission 10

Exit ticket and certificate

Correct all five items and complete the other nine missions to finish Lesson22.4.

Not complete

1. Divisible by 14 guarantees divisible by 7.

2. Not divisible by 4 means a number cannot be divisible by 12.

3. LCM of 2,3,4,5,6,7,10,11,12,13,14,15:

4. With modulus 15 and at least three selected values, the only possible shared residue is:

5. From 1 through 100, maximum size with every pair sum divisible by 10:

Exit score0 / 5

✓

Divisibility Claim & Remainder-Class Architect

This certifies that a determined mathematician completed Lesson 22.4.

Chapter 22 · Consecutive Natural Numbers

original and instructional notes

The implication explorer, exhaustive adjacent-pair checker, general claim audit, and configurable maximum-selection laboratory are added instructional scaffolds.They preserve the original conditions while making the logic visible and testable.