Chapter 22 · Consecutive Natural Numbers

Lesson 22.3 — Use Divisibility to Find Consecutive Sums

连续自然数：用整除规律寻找连续数之和

Turn a long consecutive sum into one starting value, one middle value, or one remainder condition. Then verify the actual sequence so that a necessary divisibility clue becomes a complete proof.

Divisibility narrows the search; construction proves the answer.

Grade 5Consecutive sumsOdd and even lengthsRemainder conditions10 missions

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Read a worked example, predict what the lab will show, then test your prediction. Finish a number entry and press Tab (or choose another control) to update a lab. The local hints and explanations work without opening another website.

Useful words and symbols

Consecutive means following in order. Consecutive integers have gap 1; consecutive even or odd numbers have gap 2. Positive integers start at 1. A lab explicitly allowing zero includes 0 as well.

Remainder is what is left after making equal groups. “17 mod 5 = 2” and “17 ≡ 2 (mod 5)” both say that 17 leaves remainder 2 when divided by 5. A residue class groups numbers with the same remainder.

Coprime numbers have no common factor greater than 1. The LCM is the least positive common multiple. A self-inverse remainder pairs with itself to make a multiple: for example 5+5 is a multiple of 10. Lexicographic order compares the first differing digit, as dictionary order compares letters.

To prove an answer is the best possible, explain why a better one cannot work, then give an example that reaches your bound.

Mission 1

Build every consecutive sum from one starting value

If the first term is a and there are k consecutive positive integers, every term and the entire sum are already determined.

Not complete

The sequencea, a+1, …, a+k−1

The sumS = ka + k(k−1)/2

The averageS/k = (first+last)/2

Consecutive-sum builder

First termNumber of terms

Checkpoint — use these fixed questions, even if you changed the explorer

Middle of 12,13,14,15,16Sum of 12 through 16Sum of 9,10,11,12

Mission 2

Odd lengths — the middle number carries the whole sum

With an odd number of equally spaced terms, every outer pair has average equal to the middle term.

Not complete

Pair around the middle

m−2m−1mm+1m+2

5 terms → S = 5m

The same idea works for 3, 7, 9, 11, or any odd number of terms.

Odd-length signature

k odd → S ≡ 0 (mod k)

The sum must be divisible by the number of terms, and the quotient is the middle term.

Divisibility gives the middle. You must still check that the first term is positive when the problem asks for positive natural numbers.

Odd-length representation explorer

Target sumOdd number of terms

Checkpoint — use these fixed questions, even if you changed the explorer

Middle of three consecutive numbers totaling 117First of five consecutive numbers totaling 750Last of that five-term sequence

Mission 3

Even lengths — use equal first–last pairs

An even-length sequence has no single middle term. Pair the first with the last, the second with the second-last, and so on.

Not complete

Four consecutive integers

a+a+32a+3

a+1+a+22a+3

S = 2(2a+3) = 4a+6

Therefore S ≡ 2 (mod 4)

General even-length signature

k even → S ≡ k/2 (mod k)

There are k/2 equal pairs, and each pair sum is odd.

10 terms → S ≡ 5 (mod 10)

20 terms → S ≡ 10 (mod 20)

Even-length representation explorer

Target sumEven number of terms

Checkpoint — use these fixed questions, even if you changed the explorer

First of four consecutive numbers totaling 54Last of that sequenceLargest of 20 consecutive positive integers totaling 290Required remainder modulo 10 for any 10-term sum

Mission 4

Worked example 2 — combine two remainder conditions

Three consecutive positive whole numbers have a sum divisible by 13. Their greatest member leaves remainder 4 when divided by 9. Find the least possible triple.

Not complete

Example 2 · pages 127–128

Let the middle number be a

a−1, a, a+1

Because 3 and 13 have no common factor, if 13 divides 3a, then 13 must divide a.

Sum condition3a is divisible by 13a ≡ 0 (mod 13)

Largest-number conditiona+1 ≡ 4 (mod 9)a ≡ 3 (mod 9)

First match: a = 39 → 38, 39, 40

The sequence is least because 13 and 26 fail the modulo-9 condition, while 39 is the first multiple of 13 that succeeds.

Checkpoint — use these fixed questions, even if you changed the explorer

Least middle value aSmallest numberGreatest numberSum of the triple

Mission 5

Verify whether a target really is a consecutive sum

A divisibility pattern is useful only if the calculated first term is a permitted whole number.

Not complete

General representation verifier

Target sum SNumber of terms kSmallest allowed first term

Checkpoint — use these fixed questions, even if you changed the explorer

First of five consecutive positive integers totaling 105Last of that sequenceCan 100 be four consecutive positive integers?Can 45 be ten consecutive positive integers?

Mission 6

Guided Practice 2 — satisfy lengths 3, 4, and 5 at once

Find every number from 700 through 1000 that can be written as a sum in each of three separate ways: using exactly 3, exactly 4, and exactly 5 consecutive positive whole numbers. Include both endpoints of 700–1000.

Not complete

Guided Practice 2 · page 128

Combine the three signatures

3 termsN ≡ 0 (mod 3)middle = N÷3

4 termsN ≡ 2 (mod 4)stronger than merely “even”

5 termsN ≡ 0 (mod 5)middle = N÷5

CombinedN ≡ 30 (mod 60)one candidate every 60

The worked example hint notes divisibility by 3, 2, and 5. The four-term condition must be sharpened to N ≡ 2 (mod 4), then each candidate must be constructed.

Inspect one qualifying total

Choose a target

Checkpoint — use these fixed questions, even if you changed the explorer

Number of qualifying totalsFirst qualifying totalStep between successive totalsLast qualifying total

Mission 7

original Exercise 10 — one total, three different lengths

Find the least positive natural number that can be written as sums of 9, 10, and 11 consecutive positive integers.

Not complete

Exercise 10 · page 130

9 termsN ≡ 0 (mod 9)odd length

10 termsN ≡ 5 (mod 10)even length

11 termsN ≡ 0 (mod 11)odd length

Search multiples of 99 for the first value ending in 5

Least total = 495

Checkpoint — use these fixed questions, even if you changed the explorer

Least totalFirst term of the 9-term sequenceFirst term of the 10-term sequenceFirst term of the 11-term sequence

Mission 8

Search several consecutive-sum conditions at once

The laboratory verifies every requested length and every positivity bound. It counts all matching totals in the inclusive range. It displays the first 200 matches when there are more. Each requested length gives a separate sum with the same total.

Not complete

Simultaneous consecutive-sum search

Lower boundUpper boundLengths, comma-separatedMinimum first term

Checkpoint — use these fixed questions, even if you changed the explorer

How many totals in 1–2000 have separate positive-integer representations of lengths 9,10,11?Second such matchLeast positive total for lengths 3 and 4How many totals in 700–1000 have separate positive-integer representations of lengths 3,4,5?

Try it on your own — a new problem

This extra practice has its own saved checkpoint. Try it, check your answer, then use a hint or the separate worked review.

What is the least positive total that is a sum of both 3 and 4 consecutive positive integers?

Your answer

Explain your reasoning: Why does passing a remainder test not always prove a positive sequence exists?

Mission 9

Independent consecutive-sum workshop

Use formulas and remainder signatures instead of listing random sequences. Correct all eight answers to complete the mission.

Not complete

1. Sum of 8,9,10,11,12

2. Middle of three consecutive integers totaling 117

3. First of four consecutive integers totaling 54

4. Greatest of the least positive consecutive triple whose sum is divisible by13 and whose greatest term leaves remainder4 on division by9

5. How many totals in700–1000 can each be written with3,4,and5 consecutive positive integers in separate sums?

6. Difference between consecutive qualifying totals in Question5

7. Least total represented by 9,10,11 consecutive positive integers

8. Largest of 20 consecutive positive integers totaling 290

Workshop score0 / 8

Mission 10

Exit ticket and certificate

Answer every item correctly and complete the other nine missions to unlock your certificate.

Not complete

1. Can 100 be written as four consecutive positive integers?

2. First of five consecutive positive integers totaling 105

3. Least positive total representable by both 3 and 4 consecutive positive integers

4. Greatest total in700–1000 with separate positive-integer representations of lengths3,4,and5

5. First term of the 10-term representation of 495

Exit score0 / 5

Optional reflection — When was divisibility only a clue, and when did it completely determine the sequence?

Chapter 22 Achievement

Consecutive-Sum Divisibility Navigator

This certifies that

Student

completed Lesson 22.3 and used odd–even length signatures, remainder conditions, and explicit constructions to solve consecutive-sum problems.

Date:

original and adaptation note

The four-term condition is stated in its stronger form, N ≡ 2 (mod 4), and every candidate is verified by constructing the actual sequence.The general verifiers and simultaneous search tools are added instructional scaffolds.