Chapter 22 · Lesson 5

Pair Terms and Recover Missing Sums

配对数项，还原缺失的和

Pair neighboring or matching positions before calculating. A fixed one-step gap, a square total, or a triangular boundary can reveal a missing subtotal without listing every term.

Pair the terms → count the fixed gap → rebuild the whole.

Worked example 5Exercises 5, 8, and 1110 interactive missionsWorks offline

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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

Match odd and even positions

Count positions from1 at the smallest term, moving upward by1. “Odd positions” means positions1,3,5,…, even when the numbers in those positions are even. In an even-length run of consecutive integers, every even-positioned term is exactly 1 greater than the odd-positioned term beside it.

Not complete

Pair the sequence

First termNumber of terms (even)

The reusable relationship

E = O + m

Here O is the odd-position sum, E is the even-position sum, and m is the number of pairs.

Other subtotalE = O + m

Whole totalT = 2O + m

Reverse formT = 2E − m

Checkpoint

Twelve consecutive integers are paired. Their odd-position subtotal is 300.

Number of pairsEven-position subtotalWhole totalIs each even-position term 1 larger?

Mission 2

Rebuild the worked example’s 2,012-term total

Start with exactly2,012 consecutive positive whole numbers in increasing order. Remove positions2,4,6,…,2012. The sum of the remaining odd-positioned terms is 1,025,114.

Not complete

original EXAMPLE 5

One fixed extra unit in each pair

There are 1,006 pairs. In every pair, the even-positioned term is 1 more than the matching odd-positioned term.

Odd-position sum1,025,114given

+ 1,006 →

Even-position sum1,026,120recovered

Whole total: 1,025,114 + 1,026,120 = 2,051,234.

Checkpoint

Number of pairsRecovered even-position sumAll 2,012 terms totalFirst term of the sequence

Mission 3

Recover the hidden subtotal and endpoints

A subtotal can reveal not only the missing half, but also the first and last terms—provided the data describe an actual consecutive sequence.

Not complete

Total number of termsKnown subtotalKnown sum

Checkpoint: 10 terms with odd-position sum 55

Even-position sumWhole totalFirst termLast term

Mission 4

Cancel a common factor in neighboring products

For three consecutive numbers, the two neighboring products share the middle factor. Factor before multiplying.

Not complete

original EXERCISE 5

The later product exceeds the earlier product by 114

(n+1)(n+2) − n(n+1) = 114

Factor the difference

(n+1)[(n+2)−n] = 2(n+1)

The product difference is twice the middle number.

middle = 114 ÷ 2 = 57

Recover the triple

56first

57middle

58last

56 + 57 + 58 = 171

Product-gap laboratory

Later product − earlier product

Checkpoint for the practice problem

Middle numberFirst numberLast numberSum of the three numbers

Mission 5

Recover one erased odd number from a square total

The first n odd numbers always total n². One erased term is therefore the gap between a nearby square and the remaining sum.

Not complete

original EXERCISE 8

One term is erased from a finite list of the first n positive odd numbers:1,3,5,…,2n−1

The surviving odd numbers total 1,998.

Next square above 1,99845² = 2,025

Erased term = 2,025 − 1,998 = 27

Why only one square can work

The first n odd numbers total n². The erased number is between 1 and 2n−1, so the remaining sum R satisfies (n−1)² ≤ R < n². Here 44² = 1936 ≤ 1998 < 2025 = 45². Only n = 45 fits these consecutive-square bounds.

If the sequence ends at its 45th odd number, the largest available term is:

2×45−1 = 89

The gap 27 is odd and lies from 1 through 89, so it is a valid erased term.

Missing-odd-term laboratory

Remaining sum

Checkpoint for the practice problem

Number of original odd termsOriginal full totalErased odd numberLargest original odd number

Mission 6

Find the boundaries of a triangular group

Groups contain 1 number, then 2 numbers, then 3, and so on. T(k)=1+2+…+k=k(k+1)÷2 counts the numbers through group k; define T(0)=0. These triangular numbers mark the endpoints.

Not complete

original EXERCISE 11

Grouping pattern

(1), (2,3), (4,5,6), (7,8,9,10), …

First in group kT(k−1)+1

Last in group kT(k)

Group sumk(first+last)/2

Triangular-group laboratory

Group number k

Checkpoint for group 40

First numberLast numberNumber of termsGroup sum

Mission 7

Fold arithmetic progressions into equal pairs

The first and last terms have the same pair sum as the second and second-last. This makes long progressions short.

Not complete

TEST 22 · QUESTION 1

Factor out the repeated 99

99+198+…+990 = 99(1+2+…+10)

99×55 = 5,445

TEST 22 · QUESTION 9

Average the endpoints

The multiples of 3 from 1 through 100 run from 3 to 99.

average = (3+99)÷2 = 51

Arithmetic-progression laboratory

First termLast termCommon difference

Checkpoint

Number of terms in 99,198,…,990First-plus-last pair sumTotal of the progressionAverage of 3,6,…,99

Mission 8

Audit a printed symmetric sum instead of silently repairing it

Compare two expressions carefully: the first skips 4 on the rising side, while the second includes every rising term. Their totals differ.

Not complete

PRINTED EXPRESSION

Read literally

1+2+3+5+…+49+50+49+…+6+5

2,486

The missing 4 lowers the total with every rising term included by 4.

REFERENCE-ANSWER INTENT

Include the missing 4

1+2+3+4+5+…+49+50+49+…+6+5

2,490

This is the value used by the worked example’s answer section.

Symmetric-sum laboratory

PeakOmit this rising value (0 = none)Descending side stops at

TEST 22 · QUESTION 14

Another fixed-gap recovery: carries

When two whole numbers with digit sums35and26 are added using standard column addition, exactly three columns produce a carry. Count chained carries separately. Each carry reduces the combined digit sum by9:

35+26−3×9 = 34

Checkpoint

Literal printed totalReference-answer intended totalDifference between themDigit sum after three carries

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.

The first several odd numbers have one term erased; the remaining sum is 93. Which term was erased?

Your answer

Explain your reasoning: Why would a remaining sum of 92 be impossible?

Mission 9

Independent workshop

Consecutive sequences here increase by 1, and positions are counted from 1 at the first term. Correct all eight answers to complete the workshop.

Not complete

1. For 18 consecutive terms, even-position sum minus odd-position sum:

2. Number of neighboring pairs in2,012 consecutive integers:

3. First of2,012 consecutive positive integers whose odd-position sum is1,025,114:

4. Total of those2,012 terms in Question3:

5. Sum of three consecutive positive integers when the later neighboring product minus the earlier is114:

6. One of the first n positive odd numbers is erased, leaving sum1,998. Which number was erased?

7. Write 1,2,3,… in groups (1), (2,3), (4,5,6), …, with group k containing k entries. Find the sum of group 40:

8. Sum of1through50 with4omitted, followed by49down through5 (both endpoints included):

Workshop score0 / 8

Mission 10

Exit ticket and certificate

Correct all five items and complete the other nine missions to finish Lesson22.5.

Not complete

1. Twenty consecutive terms have odd-position sum 500. What is the whole total?

2. Three consecutive positive integers a,a+1,a+2 satisfy (a+1)(a+2)−a(a+1)=82. Their sum is:

3. Erase one of the first n positive odd numbers, leaving total 2000. Which odd number was erased?

4. In groups (1), (2,3), (4,5,6), … of the positive integers, where group k contains k entries, find the sum of group 20:

5. Average of 4,8,12,…,100:

Exit score0 / 5

✓

Pairing & Missing-Sum Architect

This certifies that a determined mathematician completed Lesson 22.5.

Chapter 22 · Consecutive Natural Numbers

Optional reflection

Which pairing idea saved you the most calculation?

original and instructional notes

The printed Test 22 Question 5 omits 4 on the rising side, while the reference answer uses the version with every rising term included; this lesson reports both values rather than silently changing the original.