Combine Conditions and Fill Missing Digits
Lesson 15.3 · 组合整除条件与补全数字
Turn several divisibility clues into one coordinated search. Use least common multiples, ending-digit rules, digit sums, alternating sums, and careful candidate tables to recover hidden digits.
Turn several divisibility conditions into one target
A number divisible by two divisors must be a multiple of their least common multiple.
Use the LCM—not blind multiplication
If the divisors share no factor greater than 1, their LCM is their product:
But shared factors must not be counted twice:
A number passing both tests is a multiple of the LCM.
Find the first common multiple at or above the starting number
For a starting number N and target L:
This tells us how to complete missing ending digits or how much to add.
LCM and next-multiple laboratory
Enter whole numbers up to 9007199254740991. Both divisors must be positive; the starting number may be zero.
Check the planning step
Worked example: complete 2857□□
The six-digit number begins with 2857 and must be divisible by both 11 and 13. Each box is one digit from 0 through 9; the two boxes may differ. Run the complete ending search before checking this mission.
Replace two tests with one multiple of 143
Test the final two digits
Record the worked solution
Guided practice: add the smallest amount to 1,996
Find the smallest nonnegative whole number to add; the result must be divisible by both 23 and 19.
Build the common target
The next multiple is one more group of 437.
Subtract the remainder from one full group
Move along the next-multiple track
Complete the guided practice
Use the strongest condition first: make 9a8a divisible by 15
Because 15=3×5, the same digit a must pass both the last-digit test and the digit-sum test.
Test for 5 first
The last digit is a. A multiple of 5 must end in 0 or 5:
This reduces ten candidates to two before we use the test for 3.
Then test the digit sum
If a=0, the sum is 17. If a=5, the sum is 27.
Candidate explorer
| a | Number | Test for 5 | Digit sum | Test for 3 | Survives? |
|---|
Finish the worked example
Find only what the problem asks: a+b+c
The seven-digit number 20a0b9c is divisible by 33. The worked example asks for the sum of the three missing digits, not the digits individually.
Let:
Condition 1 · Divisible by 3
So 11+S must be divisible by 3.
Condition 2 · Divisible by 11
So S−7 must be divisible by 11.
Test every possible sum S from 0 through 27
| S | 11+S divisible by 3? | S−7 divisible by 11? | Result |
|---|
Record the invariant sum
Guided practice: complete A1999311B for divisibility by 72
Use 72=8×9. The last-three-digit test fixes B before the digit-sum test fixes A.
Find B with the test for 8
The last three digits are:
Test 110 through 119. Only 112 is divisible by 8, so B is determined.
Find A with the test for 9
After B is known, the digit sum is:
The leading digit A is from 1 through 9.
Two-stage candidate tables
Stage 1 · B candidates
| B | Last three digits | Divisible by 8? |
|---|
Stage 2 · A candidates after B=2
| A | Digit sum A+35 | Divisible by 9? |
|---|
Complete the guided practice
Use a general missing-digit search laboratory
Letters stand for digits from 0 through 9. Repeated letters must use the same digit every time; different letters may share a digit unless a problem says otherwise. Adjacent digits and letters form a number, not a product.
Pattern rules
- Use digits and capital letters only, such as 9A8A or 20A0B9C.
- The first digit cannot be zero.
- Enter one or more divisors separated by commas.
- Use 1–12 characters with one to three different letters. Divisors must be positive whole numbers no larger than 9007199254740991; commas separate divisors, not digit groups. The laboratory checks every permitted assignment.
Pattern solver
Investigate the worked example pattern 20A0B9C
Load the preset, run the search, then answer.
Let one clue fix a digit before using the others
Treat these as two separate problems: x and y are digits from 0 through 9 in each problem. The notation y6 means 10y+6, not multiplication. Run the candidate search to inspect both results before checking.
2x3y6 is divisible by 72, and y6 is divisible by 9
First use the two-digit condition:
Then the last three digits are 336, which pass the test for 8. Finally:
The digit sum must be divisible by 9.
2x36y is divisible by 55
Because 55=5×11:
- y must be 0 or 5;
- the alternating difference is:
Test the two possible ending digits and keep the case that makes the difference a multiple of 11.
Record both results
Combined-condition workshop
Solve all eight. Correct all eight responses to complete the workshop.
Worked explanation for question 1
Shared factors count once: multiples first meet at 12.
Worked explanation for question 2
100 = 12 × 8 + 4; add 12 − 4.
Worked explanation for question 3
The target is 77; 1200 + 32 = 1232 = 77 × 16.
Worked explanation for question 4
The final 5 passes 5; 4+a+5=9+a passes 3 for a=0,3,6,9. This question asks for the smallest possible a.
Worked explanation for question 5
Last two digits 04 pass 4 and digit sum 6 passes 3.
Worked explanation for question 6
11+a must be a multiple of 9; only a=7 works.
Worked explanation for question 7
8 and 15 are coprime, so multiply.
Worked explanation for question 8
240 is already a multiple of 120.
Objective exit ticket
Complete all ten missions, including a perfect exit score, to earn the certificate. Every item has one objective response.
Worked explanation for question 1
LCM=36 and 252=36×7.
Worked explanation for question 2
The next multiple of 77 is 3465.
Worked explanation for question 3
11+a must pass 3; the smallest possible sum is 12.
Worked explanation for question 4
9+a can be 9 or 18, so a can be 0 or 9. The smallest is 0.
Worked explanation for question 5
6 and 15 share factor 3; 6×15÷3=30.
Certificate of mathematical thinking
Missing-Digit Condition Solver
has completed Lesson 15.3 by coordinating several divisibility tests and proving which digit candidates survive.
Why does the order of the tests matter?
Choose one practice problem and explain why beginning with the last-digit or last-three-digit condition creates a shorter search than beginning with every possible digit assignment.
About this lesson
The LCM laboratory, exhaustive candidate tables, general pattern solver, workshop, and exit ticket are added instructional scaffolds. The lesson preserves the worked example’s requested quantities: in particular, Example 4 asks only for a+b+c, because the individual digits are not uniquely determined.