Group Numbers to Force a Relationship
The relationship you want determines the drawers you should build. Pair neighbors to force coprime numbers, pair distant values to force a fixed gap, assign factor drawers to force a common divisor, and use residue classes to control sums and differences.
Do not choose drawers before you know what must be forced.
Coprime? Fixed difference? Shared divisor? Forbidden pair sum?
The drawers may be pairs, factor families, or residue classes.
Give a construction and an upper-bound argument.
The core questions
Let the desired relationship choose the drawer design
Two consecutive numbers have GCD 1.
A completed pair has a fixed gap.
Two values in one remainder drawer differ by a multiple of 7.
Assign each number to one divisor it has.
Drawer-design checkpoint
Need a hint? Start here
Ask what a pair in the same drawer must share.
Show the reasoning
Use fixed-gap pairs for an exact difference, remainder classes for divisibility of a difference, consecutive pairs for coprimality, and factor families for a shared divisor.
Worked example 1A — pair neighbors to force coprime numbers
Fifty consecutive-pair drawers
Interactive pair board
Checkpoint
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Pair 1 with 2, then 3 with 4, and continue.
Show the reasoning
There are 50 consecutive pairs. Choosing 51 numbers forces both from one pair. Consecutive numbers have no common divisor greater than 1.
Worked example 1B — pair distant values to force a fixed gap
Pair n with n+50
Fixed-gap pair board
Checkpoint
Need a hint? Start here
Pair each small number with the number 50 above it.
Show the reasoning
The 50 drawers are {1,51}, {2,52}, …, {50,100}. A 51st choice forces a complete pair with difference 50.
Worked example 1C — force eight numbers with a shared divisor
Drawer 2
Assigned numbers divisible by 2.
Drawer 3
Assigned numbers divisible by 3.
Drawer 5
Assigned numbers divisible by 5.
Drawer 7
Assigned numbers divisible by 7.
Exceptional set: 22
Worst-case count
Most even four-drawer distribution
Trace actual numbers into factor drawers
Assign each eligible number to its smallest divisor among 2, 3, 5, 7. Thus 30 belongs only to drawer 2; 45 belongs only to drawer 3; 35 belongs only to drawer 5; 49 belongs to drawer 7. These are real assignments, unlike the balanced bars, which illustrate just the counting bound.
Drawer 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100
Drawer 3: 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99
Drawer 5: 5, 25, 35, 55, 65, 85, 95
Drawer 7: 7, 49, 77, 91
Checkpoint
Need a hint? Start here
Separate numbers with no divisor among 2,3,5,7.
Show the reasoning
The exceptions are 1 and the 21 primes above 7. At least 51−22=29 choices enter four factor drawers, using one assignment per number. Since 4×7=28, at least 8 share a divisor.
Build fixed-gap pairs and same-remainder drawers
Fixed gap in a doubled range
In 1,2,…,2d, pair i with i+d.
Difference divisible by a modulus
Put distinct integers into drawers by their remainder modulo m.
Checkpoint
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Make the intended relationship automatic inside each drawer.
Show the reasoning
For gap 40 in 1–80, the 40 pairs {i,i+40} force a pair at 41 choices. For difference divisible by 7, there are seven remainders, so eight integers force a match.
Guided Practice 1 — build the largest set with no pair sum divisible by 7
Remainder counts in 1 through 50
Construct a safe set
At most one residue-0 value is safe.
Checkpoint
What is the greatest possible size of a subset of 1,2,…,50 in which no two distinct selected numbers have a sum divisible by 7?
Need a hint? Start here
Pair remainder classes that add to 7.
Show the reasoning
Choose one class from each pair (1,6), (2,5), (3,4), taking the larger available class. Allow at most one remainder-0 value. The counts give 8+7+7+1=23, and this construction reaches the upper bound.
Use a general complementary-residue laboratory
Maximum safe subset search
Residue counts and chosen classes
Checkpoint
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Self-complementary remainders need special care.
Show the reasoning
For modulus m, pair r with m−r. Take the larger class, never both; a self-complementary class allows at most one value. This gives 23 for 1–50 modulo 7 and 16 for 1–30 modulo 9.
Force a same-remainder pair or a three-number sum
Difference divisible by 7
Enter2–100distinct integers to explore. With at least eight, a match is guaranteed. Two must have the same remainder modulo 7.
Three numbers whose sum is divisible by 3
Enter exactly five integer entries; equal values in different positions are allowed. Choose three different entries, not necessarily three different values. Either one residue class contains three values, or all three residue classes occur.
Checkpoint
Need a hint? Start here
For triples use the three remainder classes modulo 3.
Show the reasoning
Five integers either give three of one remainder, whose sum is divisible by 3, or all three remainders appear and you can select 0,1,2. For differences, eight integers force two of the same remainder modulo 7.
Try a fresh problem
What is the least number of integers that guarantees two have a difference divisible by 11?
This fresh problem has its own checkpoint. Your written explanation is saved for comparison and is not automatically graded.
Worked explanation for the fresh problem
There are eleven remainders modulo 11. Twelve integers force two of the same remainder. Eleven can fail: choose 0 through 10.
Relationship-drawer workshop
Need a hint? Start here
Revisit the mission connected to each question.
Show the reasoning
Answers:50,50,22,8,23,16,8,5. In the factor question,8is the bound from rounding29÷4upward. For pair-sum avoidance, compare complementary remainder classes and permit only one from each self-complementary class.
Exit ticket and certificate
Relationship Drawer Architect
This certificate recognizes
for grouping numbers into pairs, factor families, and residue classes to force precise mathematical relationships.
Lesson 23.2 completed
Need a hint? Start here
Solve each item without using the answer shown in an earlier example.
Show the reasoning
Answers:31,8,14,12,5. For1–36 modulo6, the two complementary pairs contribute6each, and residues0and3permit1each:14. Four entries1,1,2,2have no triple divisible by3, proving five are needed.
original and lesson scope
This lesson develops Chapter 23 Example 1 and Guided Practice 1 from this lesson, together with the chapter exercises about sums divisible by 3 and differences divisible by 7. The modulo-9 maximum-selection problem is transferred from Test 23. The interactive exhaustive searches and general laboratories are added learning supports.