15.5Grade 5 Math Lab
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Chapter 15 · Divisibility and remainder classes

Track Repeated Elimination with Remainder Classes

Lesson 15.5 · 用余数类追踪重复淘汰

A circle with thousands of pieces looks impossible to simulate. Number the positions, preserve the counting phase, and turn repeated “skip one, remove one” moves into simple remainder patterns.

Number the positions. Follow the phase. Count the surviving class.
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Mission 1

Turn the moving rule into numbered positions

The pieces move nowhere; only our attention moves. Numbering the positions turns a physical process into arithmetic.

Not complete

The exact rule

Begin at the black piece. Move clockwise. Skip the next living piece, remove the following living piece, and continue without restarting the count after a full circuit.

black startskip 1remove 2skip 3remove 4

Removed pieces no longer count when skipping or removing. Keep the original labels; never renumber survivors. Stop immediately when black is removed. One removal is one operation; skips do not count as removal operations. If black is the only piece left, continue the same rule: skip it once, then remove it.

Number a 13-piece model

Place 12 white pieces after the black piece, moving clockwise. Number the white pieces:

1, 2, 3, …, 12

Give the black piece the next label:

black = 13

This small model has the same structure as the practice problem, but every position can be seen.

Check the setup

Mission 2

Watch the complete 13-piece process

A small simulation reveals the phase change that is easy to miss at the end of the first circuit.

Not complete
blacklast skippedlast removedsurvivor
Last skipped
Last removed
Removal number0
White pieces alive12
Removal sequenceblue = first circuit · gold = later
No pieces removed yet.
After 12 is removed, the next living piece is black. Black is skipped, so white piece 1 is removed next. The count does not restart.

Read the completed model

Mission 3

Apply the first circuit to 2,013 positions

Now use the complete practice problem. The ring contains one black Go piece and 2,012 white Go pieces.

Not complete
original problem · Chapter 15, Example 6

One black piece and 2,012 white pieces

Number the white pieces from 1 through 2,012 clockwise, beginning with the first white piece after black. Number the black piece 2,013. Starting from black, skip one living piece and remove the next. How many white pieces remain when black is removed?

First circuit: remove the evens

2, 4, 6, …, 2012

These are precisely the positions with:

n ≡ 0 (mod 2)

The count is:

2012 ÷ 2 = 1006

What remains after the first circuit?

1, 3, 5, …, 2011, 2013

There are 1,006 odd-numbered white pieces plus the black piece:

1006 + 1 = 1007 pieces alive

Reveal the first-circuit ledger

White pieces at start2,012
Black label2,013
First-circuit removals?
Total pieces still alive?

Check the first circuit

Mission 4

Keep the counting phase through the second circuit

The key move occurs after white piece 2,012 is removed: black is the next living piece, so black is skipped and white piece 1 is removed.

Not complete
remove 2012skip black 2013remove 1skip 3remove 5

Second-circuit removals

1, 5, 9, …, 2013

They differ by 4 and all satisfy:

n ≡ 1 (mod 4)

The black piece is included because:

2013 = 4 × 503 + 1

White positions still alive

3, 7, 11, …, 2011

They differ by 4 and all satisfy:

n ≡ 3 (mod 4)

Every even position was removed earlier.

Four remainder classes modulo 4

0 mod 4

4, 8, 12, …

1 mod 4

1, 5, 9, …

2 mod 4

2, 6, 10, …

3 mod 4

3, 7, 11, …

Check the second circuit

Mission 5

Count the surviving arithmetic sequence

Knowing the survivor class is not enough. Count every term from the first survivor through the last, including both endpoints.

Not complete

Direct count

3, 7, 11, …, 2011
last − first2011 − 3
divide by step2008 ÷ 4
include the first term502 + 1
survivors503

Second ledger check

After the first circuit, 1,007 pieces are alive. The second-circuit removal sequence:

1, 5, 9, …, 2013

contains:

(2013 − 1) ÷ 4 + 1 = 504

pieces, including black. Thus:

1007 − 504 = 503

Arithmetic-sequence counter

Enter nonnegative whole-number first and last terms with last ≥ first, and a positive whole-number step. The last term must be reached exactly. For this mission, run the counter with 3, 2011, and step 4.

Difference
Number of steps
Number of terms
Exact fit?

Complete both counts

Mission 6

Read a position from its remainder class

The notation n ≡ r (mod 4) means that n leaves remainder r when divided by 4.

Not complete

Preview of original-ring positions 1 through 24 (not a separate 24-piece ring)

Choose all three views. The final view classifies every position by the circuit that removes it.

3 mod 4

Positions such as 3, 7, 11, 15, 19, 23, … survive until black is removed.

1 mod 4

Positions such as 1, 5, 9, 13, 17, 21, … are removed during the second circuit.

0 or 2 mod 4

These are the even positions, removed during the first circuit.

Classify these positions in the worked example ring of 2,013 pieces

Mission 7

Build a shortcut for 4m white pieces

The worked example numbers are not accidental: 2,012 = 4 × 503. The same two-circuit structure works for W=4m white pieces, where m is a positive whole number. Every survivor count in this lesson is measured at the instant black is removed, not after eliminating every piece.

Not complete
2012
W = 4m, so total positions = 4m + 1
m503
First-circuit removals1006
Second removals incl. black504
White survivors503

Why the formula works

If there are 4m white pieces, the black label is 4m+1.

  • First circuit removes 2m even positions.
  • Second circuit removes m+1 positions of the form 4k+1, including black.
  • The remaining white positions are 3, 7, …, 4m−1: exactly m pieces.

original as one family member

2012 = 4 × 503
2013 = 4 × 503 + 1
white survivors = 503

Use the family rule

Mission 8

Use a simulator when the shortcut does not fit

The W÷4 shortcut is proved only when W is a multiple of 4. For other ring sizes, simulate the same skip/remove rule and keep the phase exact.

Not complete

General ring simulator

Run the simulator to display the ring at the instant black is removed.

Total positions
Black removed on operation
White pieces left
W÷4 shortcut
Removal sequence through black
No simulation yet.
Surviving white labels will appear here.

Run both the 10-white and 6-white simulations, then compare

Mission 9

Remainder-class workshop

Answer all eight questions. Correct all eight responses to complete the workshop.

Not complete
Worked explanation for question 1

White labels are 1–28, so black is 29.

Worked explanation for question 2

Remove the 14 even labels 2 through 28.

Worked explanation for question 3

28=4×7, so the family rule gives 7.

Worked explanation for question 4

The family has m+1=8 second removals.

Worked explanation for question 5

(59−3)÷4+1=15.

Worked explanation for question 6

87=4×21+3; it belongs to the survivor class in the worked example ring.

Worked explanation for question 7

Skip black 29, then remove white 1.

Worked explanation for question 8

124÷4=31.

Workshop score0 / 8
Mission 10

Objective exit ticket

Complete all ten missions, including a perfect exit score, to earn the certificate.

Not complete
Worked explanation for question 1

36=4×9.

Worked explanation for question 2

m+1=9+1=10.

Worked explanation for question 3

(99−3)÷4+1=25.

Worked explanation for question 4

101=4×25+1; the black label of a 100-white ring is removed second circuit.

Worked explanation for question 5

236 is divisible by 4, so 236÷4=59.

Exit score0 / 5

Certificate of mathematical thinking

Remainder-Class Elimination Navigator

Grade 5 Mathematician

has completed Lesson 15.5 by translating a circular elimination process into numbered positions, residue classes, and exact survivor counts.

Optional reflection

Why must the count continue across the starting point?

Explain what would go wrong if a student restarted “skip one, remove one” after each circuit instead of preserving the phase.

Learning notes

About this lesson

The 13-piece simulator, the explicit remainder-class diagrams, the 4m family proof, the arbitrary-size simulator, workshop, and exit ticket are added instructional scaffolds. They preserve the worked example rule and make the counting phase visible.