Grade 5 · Chapter 12 · Lesson 3

Discover Hidden Periods from Equal Window Sums

相等窗口和中的隐藏周期

A sequence may repeat even when nobody tells us the repeating block. We will slide equal-length windows, cancel their shared terms, and prove which far-apart positions must match.

Same total + same overlap → the leaving value equals the entering value.

10 interactive missionsTwo complete original solutionsEverything needed is on this page.Objective checking

Mission 1

Slide a window across a sequence

A window is a fixed number of consecutive positions viewed together. This sample arrangement repeats the five values 2, 3, 1, 4, 4, so every five-box window contains the same five values, possibly in a different order and has total 14.

Not complete

Window begins at Box 1

Highlighted positions: 1–5·window sum = 14·Move one step: four boxes stay, one leaves, one enters.

Read the model

How many boxes are in one window?

After moving one step, how many boxes overlap?

What is the sample window sum?

Mission 2

Cancel the shared overlap

Now use symbols. The first two five-box windows have equal totals. Four terms occur in both equations, so they can be removed from both sides.

Not complete

Reveal the cancellation

Stage 1: compare both equal totals.

Why this works: if two totals are equal and they contain the same four middle terms, the only remaining terms must also be equal.

Which term leaves when the window moves?

Which term enters?

If x₁ = 2, what is x₆?

Mission 3

Build the general hidden-period rule

If every window has length k and every window sum is equal, comparing windows whose starting positions differ by one proves that positions k places apart have equal values.

Not complete

Window length k = 5

xᵢ = xᵢ₊₅

Precise language: positions k places apart have equal values wherever both positions exist in the row. If this pattern continues indefinitely, k is a repeat step, but the smallest period may be smaller.

Equal 4-box window sums imply x₃ = x_?.

Equal 6-box window sums imply x₂ = x_?.

If every 4-box window has the same sum, must 4 be the smallest period?

Mission 4

Follow the worked example’s 36-box equality chain

Complete problem: 100 small balls are placed in 36 boxes in a row. Every five consecutive boxes contain 14 balls in total. Box 1 contains 2 balls. Find the number in Box 36.

Not complete

Move five positions at a time

Equal five-box window sums give:

x₁ = x₆ = x₁₁ = x₁₆ = x₂₁ = x₂₆ = x₃₁ = x₃₆

Begin at Box 1. Reveal each guaranteed equal position.

How many jumps of 5 go from Box 1 to Box 36?

How many balls are in Box 36?

Mission 5

Use quotient and remainder to identify the class

An index is a position number. A class groups indices that give the same remainder when divided by 5. A jump of five preserves a box’s position inside a five-box cycle. Dividing the box number by 5 identifies its class.

Not complete

Box 36

36 = 7 × 5 + 1

The remainder 1 places Box 36 in the same class as Boxes 1, 6, 11, 16, 21, 26, and 31.

Class 1 modulo 5

Every highlighted index has the form 5n + 1, where n is a whole number from 0 through 7 here. Class 5 means remainder 0, the last position of a cycle.

Quotient q

Remainder r

Position inside the five-box cycle

Value in Box 36

Mission 6

Solve the same practice problem by totals

A second solution uses the total number of balls. The first 35 boxes can be split into seven non-overlapping groups of five. Every such group totals 14.

Not complete

First 35 boxes: 7 × 14 = 98

Box 36: 100 − 98 = 2

Method A — repeat classes

Proves Box 36 has the same value as Box 1.

Method B — total subtraction

Finds the total in the first 35 boxes and subtracts from 100.

Number of complete five-box groups

Balls in the first 35 boxes

Balls left for Box 36

Do both methods agree?

Mission 7

Hidden-period laboratory

Using only equal window sums and one known term, we can transfer that value to positions a multiple of the window length away. Other classes need another clue. Enter whole numbers in this laboratory; indices are position numbers starting at 1. The map previews at most the first 36 positions, while the calculation uses your full indices.

Not complete

Equal window length k

Known index

Known value

Target index

35index difference

0difference ÷ k remainder

1known residue class

1target residue class

The target is in the same class, so its value is 2.

Record three conclusions

Equal 4-box sums, x₂ = 7. Find x₁₈.

Equal 6-box sums, x₅ = 9. Find x₄₁.

Equal 5-box sums, x₁ = 2. Can x₃₅ be determined from that clue alone?

Mission 8

Hidden-period error detective

Judge each statement precisely. “Repeats every k positions” does not mean every term is equal, and k need not be the smallest period.

Not complete

A smaller period can hide inside a larger one

1, 2, 1, 2, 1, 2, …

Every four consecutive terms sum to 6, so a four-step repeat is valid. But the smallest period is 2.

One known class is not every class

For five-box windows, knowing x₁ tells us x₆, x₁₁, …, but it does not determine x₂, x₃, x₄, or x₅.

1. Equal five-box window sums imply x₄ = x₉.

2. Equal five-box window sums imply x₁ = x₂.

3. In the practice problem, x₁ = 2 is enough by itself to determine x₃₅.

4. A sequence may repeat every four positions even though its smallest period is two.

5. In the worked example’s total method, 100 − 7 × 14 gives the number in Box 36.

original Exercise 4: a hidden divisibility pattern

How many fractions in 1/6, 2/7, 3/8, …, 2009/2014 are already in simplest form? Fraction number k is k/(k+5).

Hint: compare numerator and denominator

Any common divisor of k and k+5 must divide their difference, 5. Since 5 is prime, a fraction can reduce only when k is a multiple of 5.

Worked solution

In every five consecutive numerators, exactly one is divisible by 5. There are ⌊2009 ÷ 5⌋ = 401 such numerators. Of 2,009 fractions, 2,009 − 401 = 1,608 are already in simplest form. Here ⌊2009 ÷ 5⌋ means the whole-number quotient.

Mission 9

Independent workshop

Try each new problem before opening a hint. Correct all 8 answers to complete this mission.

Not complete

1. Every 4-term window has the same sum. Term 3 is 6. Find term 31.

Hint

Subtract the indices and compare the difference with the window length.

Worked explanation

31 − 3 = 28, a multiple of 4. Equal overlapping sums force terms 4 places apart to match, so the answer is 6.

2. Every 7-term window has the same sum. Term 2 is 8. Find term 44.

Hint

Subtract the indices and compare the difference with the window length.

Worked explanation

44 − 2 = 42, a multiple of 7. Equal overlapping sums force terms 7 places apart to match, so the answer is 8.

3. Every 3-term window has the same sum. Term 5 is 9. Find term 29.

Hint

Subtract the indices and compare the difference with the window length.

Worked explanation

29 − 5 = 24, a multiple of 3. Equal overlapping sums force terms 3 places apart to match, so the answer is 9.

4. Every 4-term window totals 18. The first 21 terms total 97. Find term 21.

Hint

Group the first 20 terms into nonoverlapping groups of four.

Worked explanation

Five groups total 5 × 18 = 90. The last term is 97 − 90 = 7.

5. Equal 6-term sums are known, and term 1 is 3. Can term 8 be determined? Write yes or no.

Hint

Compare the two indices modulo 6.

Worked explanation

8 − 1 = 7 is not divisible by 6; the two terms belong to different classes.

6. Smallest period of 3, 5, 3, 5, …, even though every 4-term window totals 16?

Hint

A guaranteed repeat step may contain a shorter repeating block.

Worked explanation

The block 3,5 repeats. Its shortest length is 2.

7. Every 5-term window has the same sum. Term 4 is 7. Find term 49.

Hint

Subtract the indices and compare the difference with the window length.

Worked explanation

49 − 4 = 45, a multiple of 5. Equal overlapping sums force terms 5 places apart to match, so the answer is 7.

8. Two 8-term windows start one position apart and have the same sum. If the leaving term is 12, what is the entering term?

Hint

Cancel the seven shared terms.

Worked explanation

Equal totals minus the same seven middle terms leave equal entering and leaving terms: 12.

Mission 10

Transfer exit ticket

Try each new problem before opening a hint. Correct all 5 answers to complete this mission.

Not complete

1. Every 6-term window has the same sum. Term 4 is 11. Find term 46.

Hint

Subtract the indices and compare the difference with the window length.

Worked explanation

46 − 4 = 42, a multiple of 6. Equal overlapping sums force terms 6 places apart to match, so the answer is 11.

2. Every 3-term window totals 17. The first 16 terms total 91. Find term 16.

Hint

Split the first 15 terms into five groups of three.

Worked explanation

91 − 5 × 17 = 6.

3. Every 5-term window has an equal sum. Which term from 1–5 must equal term 47?

Hint

Use the remainder of 47 divided by 5.

Worked explanation

47 = 9 × 5 + 2, so term 47 matches term 2.

4. Equal 4-term sums and term 2 = 7 are given. Can term 20 be determined? Write yes or no.

Hint

Check whether the index difference is divisible by four.

Worked explanation

20 − 2 = 18, which is not a multiple of 4.

5. Equal 7-term window sums imply term 6 equals the next term with which index?

Hint

Add the window length to the index.

Worked explanation

6 + 7 = 13.

Optional reflection — not automatically graded

Explain why equal neighboring window totals force the term leaving the window to equal the term entering it.

Lesson checkpoints completed

Hidden-Period Window Detective

This certifies that the learner can compare equal sliding-window sums, cancel shared terms, prove a repeat step, identify residue classes, and solve the 36-box problem in two different ways.

Lesson 12.3 · Grade 5 Periodic Problems

Teaching notes

The original states that 100 balls are distributed among 36 consecutive boxes, every five neighboring boxes total 14, and Box 1 contains 2.It presents two solutions: a five-step equality chain and a subtraction method using seven groups of five.This page restates the complete problem and adds the explicit adjacent-window cancellation proof, general rule, residue-class laboratory, feedback, workshop, and exit ticket.