Construct and Enumerate Divisible Numbers
Lesson 15.6 · 构造并枚举整除数
Some divisibility problems ask for one missing digit. These ask for all solutions, the largest solution, the smallest solution, or proof that a solution is impossible. Use the strongest restriction first, keep a complete candidate ledger, and stop only when every case has been accounted for.
Turn enumeration into an organized proof
A complete search is not random guessing. It defines every candidate, applies the strongest restrictions first, and records why each candidate survives or fails.
Two earlier missing-digit searches
The worked example practice begins with two constructions already developed in Lesson 15.3:
First force y=3; then the tests for 8 and 9 force x=4. Thus xy=43.
The tests for 5 and 11 leave x=4, y=5, so the number is 24,365.
Use the strongest clue first
A clue that leaves only one or two candidates should come before a broad clue.
- Divisible by 5: inspect the final digit.
- Divisible by 8: inspect the final three digits.
- Divisible by 9: use the complete digit sum.
- Several tests together: keep only their intersection.
Review the search order
Construct every number that passes several simple tests
Use final-digit restrictions before trying permutations. This turns a large-looking construction into a tiny candidate set.
Smallest four-digit common multiple
A number divisible by 3, 4, and 5 must be divisible by:
The first multiple of 60 at or above 1000 is:
60×17=1020
Use three of 0, 3, 5, 7
Form every three-digit number, without repeating a digit, that is divisible by 2, 3, and 5.
Divisibility by both 2 and 5 forces a final digit of 0. The other two digits must have a sum divisible by 3.
Complete original enumeration
| Candidate | Even? | Digit sum ÷3? | Ends 0 or 5? | Result |
|---|---|---|---|---|
| Run the enumeration. | ||||
Record the complete result
Count every completion of a partially hidden number
A counting question requires every valid filling—not merely one example.
A multiple of 75 with two unreadable digits
The five-digit number has no repeated digits:
It is divisible by 75. How many such five-digit numbers exist?
Use 75 = 3 × 25
The final two digits must be a multiple of 25 ending in 5:
Then use the digit sum to test divisibility by 3, while rejecting any repeated digit.
Pattern search
Results
Run the search to see every valid completion.
| Number | Last two | Digit sum | Distinct? | Pass? |
|---|---|---|---|---|
| No search yet. | ||||
Explain the count
Turn a digit-sum condition into a short search
The digit sum of a three-digit number is at most 27, so equations involving the digit sum often leave only a few candidates.
A number equals 18 times its digit sum
Let the digit sum be s. Then:
Because N is three-digit and s≤27, test only the possible values of s. The unique match is:
Subtract the digit sum
For a three-digit number N, the quantity:
is always divisible by 9. The result is written as:
The only multiple of 9 from 460 through 469 is 468.
Digit-sum fixed-point laboratory
For this mission, search with multiplier 18 and exactly three digits before checking. Other settings are optional explorations.
Check the two original ideas
Interpret a place-value transformation precisely
When a zero is inserted between two digits, the new number changes place value. Translate the wording into an equation before searching.
Insert one zero between the digits
Let the original two-digit number be:
After inserting 0 between its digits, it becomes:
The worked example says the new number is “larger by eight times the original,” so the increase equals 8 originals and the new number equals 9 originals:
The only digit pair is a=4, b=5.
Verify the worked example result
Insert-zero search
For this mission, run the search with new number ÷ original = 9. The increase is a separate quantity.
Matches
Run the search.
Record the transformation
Use the digits 1–9 once and maximize three multiples of 9
A maximum problem needs a valid construction and proof that no larger construction survives.
Three three-digit multiples of 9
Use each digit from 1 through 9 exactly once to make three three-digit numbers divisible by 9. Make the three numbers as large as possible.
The three digits in each number are different, so their sum is at most 9+8+7=24. To pass 9, the sum must therefore be 9 or 18. To make “as large as possible” precise, sort the three numbers from largest to smallest, then maximize the first number, then the second, then the third. A complete search compares all valid disjoint triples.
Exhaustive maximum search
Validate a proposed construction
State the maximum construction
Search downward to prove a maximum
When the target must be a multiple of several divisors, search only multiples of their LCM—not every number.
Largest five-digit number with distinct digits
The number must be divisible by:
Search five-digit multiples of 1155 downward and keep only numbers whose five digits are different.
Four consecutive numbers
Let the numbers be:
Their sum is:
Their sum lies from 400 through 440, inclusive, and is divisible by 9. Among the candidate multiples of 9, only 414 has the required remainder 2 when divided by 4.
Maximum and consecutive-number search
Top distinct-digit multiples
Consecutive-number result
No search yet.
Record both original results
Classify impossible cases and complete solution families
A proof may end with one solution, several solutions, or no solution. State exactly what the evidence supports.
Can one number be 17 times the other?
Both are positive whole numbers whose digits come only from 1, 4, 6, and 9. Digits may repeat, and the numbers may have different lengths. The final digit of 17N depends only on 7 times N's final digit.
No such pair exists.
“Magical” numbers below 130
Let N be positive. Write the ordinary digits of N immediately after the ordinary digits of any nonnegative whole number A, without adding a leading zero to N. For example, appending 25 to 7 makes 725. Let k be the number of digits of N. If the result is always divisible by N, then:
So N must divide 10k, where k is the number of digits of N.
Three consecutive numbers
Find consecutive whole numbers with all three between 100 and 200, inclusive:
- smallest divisible by 3;
- middle divisible by 5;
- largest divisible by 7.
No search yet.
Classify the worked example results
Construction-and-enumeration workshop
Answer all eight questions. Correct all eight responses to complete the workshop.
Worked explanation for question 1
LCM=20; 100 is its first three-digit multiple.
Worked explanation for question 2
204,240,402,420 all have digit sum 6 and are even.
Worked explanation for question 3
Arrange the digits in descending order; 420 passes both 2 and 3.
Worked explanation for question 4
3+a must pass 3, so a=0,3,6,9.
Worked explanation for question 5
For two digits 10a+b=2a+2b gives b=8a; only a=1,b=8. One-digit positive numbers fail; larger numbers exceed twice their digit sum.
Worked explanation for question 6
The tens digit becomes hundreds: 3×100+6.
Worked explanation for question 7
990 has repeated 9; 975 is next and has distinct digits.
Worked explanation for question 8
They must divide 10: 1,2,5.
Objective exit ticket
Complete all ten missions, including a perfect exit score, to earn the certificate.
Worked explanation for question 1
LCM=60, and 120 is the first three-digit multiple.
Worked explanation for question 2
306,360,603,630 each have digit sum 9; leading zero is forbidden.
Worked explanation for question 3
630 has the largest hundreds digit and largest remaining tens digit.
Worked explanation for question 4
1000 is too large; 975 is the largest three-digit multiple and its digits differ.
Worked explanation for question 5
It must divide 100. Two-digit divisors are 10,20,25,50; largest is 50.
Certificate of mathematical thinking
Divisibility Construction & Enumeration Architect
has completed Lesson 15.6 by defining candidate sets, applying divisibility filters, proving maxima, and distinguishing unique, multiple, and impossible outcomes.
Why is one working example not enough?
Explain what extra work is needed when a problem asks for all solutions, the largest solution, or proof that no solution exists.
About this lesson
The three-digit number equal to 18 times its digit sum comes from Guided Practice 5 on. The thirteen Chapter 15 practice questions on provide the missing-digit reviews, the smallest common multiple, constructions using 0, 3, 5, and 7, the 3□6□5 counting problem, the inserted-zero problem, the 1–9 maximum construction, the digit-sum subtraction, the five-digit maximum, the two consecutive-number investigations, the 17-times impossibility question, and the magical-number definition.
The complete candidate ledgers, exhaustive searches, maximum audits, generalized laboratories, workshop, and exit ticket are added instructional scaffolds. Exercises 1 and 2 are briefly reviewed here because they were taught in full in Lesson 15.3.