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5.1 Is This a Multiple of? — Sums of Consecutive Numbers

🎓 Class 8 Mathematics CBSE Theory Ch 5 — Number Play ⏱ ~20 min
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Targeting Class 8 level in Number Theory, with Basic difficulty.

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5.1 Is This a Multiple of? — Sums of Consecutive Numbers

Anshu is exploring sums of consecutive numbers?. He writes:

Anshu's List

7 = 3 + 4
10 = 1 + 2 + 3 + 4
12 = 3 + 4 + 5
15 = 7 + 8
21 = 1 + 2 + 3 + 4 + 5 + 6

He wonders:

  • "Can I write every natural number as a sum of consecutive numbers?"
  • "Which numbers can I write as the sum of consecutive numbers in more than one way?"
  • "Oh, I know all small numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?"
  • "Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers."
Explore these questions yourself. Take any 4 consecutive numbers, e.g. 3, 4, 5, 6. Place '+' or '–' signs between them in all possible ways. How many different possibilities are there? Write all of them.

Eight such expressions are possible. The tree diagram below helps list them systematically:

3 +4−4 +5−5 +5−5 +6 → 3+4+5+6=18 −6 → 3+4+5−6=6 +6 → 3+4−5+6=8 −6 → 3+4−5−6=−4 +6 → 3−4+5+6=10 −6 → 3−4+5−6=−2 +6 → 3−4−5+6=0 −6 → 3−4−5−6=−12
Binary tree of +/− combinations for 3, 4, 5, 6 — 8 outcomes.
Evaluate each expression above. Notice anything interesting? All results have the same parity (all even here) — either all even or all odd!

Parity Pattern with 4 Numbers

Take any 4 consecutive numbers \(a, b, c, d\) (where \(b = a+1\), \(c = a+2\), \(d = a+3\)) and place '+' or '–' signs. Let's see why the results always have the same parity.

Explanation 1 — Algebra

Consider \(a + b - c - d\). When one sign is switched, the value changes by twice one of the numbers. Since differences of pairs are even, the parity never changes. So all 8 expressions \(\pm a \pm b \pm c \pm d\) have the same parity.

Explanation 2 — Why? (odd + even rules)

In four consecutive numbers, exactly 2 are odd and 2 are even. The parity of \(a \pm b \pm c \pm d\) depends only on the odd terms. With two odds, odd ± odd = even, so the overall expression is always the same parity.

🔵 Is there a way to explain this? Let us consider any of the 8 expressions \(a \pm b \pm c \pm d\) where \(b = a+1\), \(c = a+2\), \(d = a+3\). Replacing \(+b\) by \(-b\) changes \(b\) to \(-b\), a change of \(2b\) — always an even change. So parity is preserved.

Breaking Even

Without computing, find out which of the following arithmetic expressions have the same parity (are all even or all odd):

ExpressionParity
43 + 37even + odd = odd
672 − 88even − even = even
4 × 347 + 3even + odd = odd
708 − 477even − odd = odd
809 + 214odd + even = odd
119 + 303odd + odd = even
543 − 479odd − odd = even
\(513^2\)odd × odd = odd

Using Algebra for Letter-Numbers

Using your understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers:

(i) \(2a + 2b\)   (ii) \(3g + 5h\)   (iii) \(6m + 2n\)   (iv) \(2a - x\)   (v) \(13k - 5k\)   (vi) \(6m - 3n\)   (vii) \(x^2 + 2\)   (viii) \(h^2 + 1\)   (ix) \(4k + 3y\)
Always even: (i) 2a+2b = 2(a+b), (iii) 6m+2n = 2(3m+n), (v) 13k−5k = 8k.
Not always even: (ii) 3g+5h, (iv) 2a−x (depends on x), (vi) 6m−3n = 3(2m−n), (vii) x²+2, (viii) h²+1, (ix) 4k+3y.

Pairs to Make Fours

Take any pair of even numbers. Add them. Is the sum divisible by 4? Not always. There is a pattern: even numbers fall into two types by remainder when divided by 4.

Even, multiple of 4 (4p) Even, remainder 2 (4q+2) Pairs of 2 rows → perfectly divisible by 4 Extra 2 dots leave remainder 2
Even numbers split into two types: multiples of 4 (4p), and those leaving remainder 2 (4q+2).

Explanation with algebra

Adding two even numbers that are both multiples of 4:
\(4p + 4q = 4(p+q)\) → always a multiple of 4.

Adding two even numbers both leaving remainder 2:
\((4p+2) + (4q+2) = 4p + 4q + 4 = 4(p+q+1)\) → always a multiple of 4.

Adding a multiple of 4 and one leaving remainder 2:
\(4p + (4q+2) = 4(p+q) + 2\) → never a multiple of 4.

Activity: Sum-Explorer Game
L3 Apply
Materials: Paper, pencil, a partner
Predict: Can every natural number be written as a sum of consecutive numbers? Which numbers resist?
  1. Pick any number between 1 and 50.
  2. Try to express it as a sum of 2, 3, 4, or more consecutive natural numbers.
  3. Write how many different ways you can express it. For example: \(15 = 7+8 = 4+5+6 = 1+2+3+4+5\).
  4. Tabulate your findings for 10 numbers and look for a pattern.

Big Insight: Only powers of 2 (1, 2, 4, 8, 16, 32, ...) cannot be expressed as a sum of two or more consecutive natural numbers. Every other number can. The number of ways equals the count of odd factors greater than 1.

Always, Sometimes, or Never

We examine different factors and multiples and determine whether a statement is "Always True", "Sometimes True", or "Never True".

Statement 1: If 8 exactly divides two numbers separately, it must exactly divide their sum.
Always True. If \(a = 8p\) and \(b = 8q\), then \(a+b = 8(p+q)\) — divisible by 8. Examples: 8+16=24, 80+120=200, both multiples of 8.
Statement 2: If a number is divisible by 8, then it is also divisible by any multiple of 8.
Sometimes True. 72 is divisible by 8, and also by 24 (= 8×3). But 72 is not divisible by 16 (= 8×2). True only when the number contains extra factors matching the multiple.
Statement 3: If a number is divisible by 7, then all multiples of 7 will be divisible by it.
Always True. If \(k\) divides 7, then \(k\) is either 1 or 7. Both 1 and 7 divide every multiple of 7. In general: if \(a\) divides \(b\), then \(a\) divides every multiple of \(b\).
Statement 4: If a number is divisible by 12, then the number is also divisible by all the factors of 12.
Always True. Factors of 12 are 1, 2, 3, 4, 6, 12. Any multiple of 12 is automatically a multiple of each factor. General rule: if \(a\) divides \(b\), and \(c\) divides \(a\), then \(c\) divides \(b\).
Statement 5: If a number is divisible by 7, then it is divisible by any multiple of 7.
Sometimes True. 14 is divisible by 7, but not by 21 (a multiple of 7). True only when that specific multiple itself divides the number.

General rule: If \(A\) is divisible by \(k\) and \(A\) is also divisible by \(m\), then \(A\) is divisible by the LCM of \(k\) and \(m\).

Key Summary
  • If \(a\) divides \(b\), then all multiples of \(a\) are divisible by \(b\) — sometimes true.
  • If \(a\) divides \(b\) and \(b\) divides \(N\), then \(a\) divides \(N\) — always true.
  • If \(M\) and \(N\) are multiples of \(a\), then \(M+N\) and \(M-N\) are multiples of \(a\) — always true.

What Remains? Patterns in Remainders

Find a number that has a remainder of 3 when divided by 5. Write more such numbers.

Which algebraic expression(s) captures all such numbers?
(i) \(3k+5\)   (ii) \(3k-5\)   (iii) \(5k+3\)   (iv) \(5k-3\)   (v) \(5k-2\)   (vi) \(5k-7\)
The numbers 3, 8, 13, 18, 23 ... all leave remainder 3 on division by 5. Testing for \(k = 1, 2, 3, ...\):
(iii) \(5k+3\): 8, 13, 18, 23 ✓ (starting \(k=1\)) — captures numbers ≥ 8. At \(k=0\), gives 3. ✓
(v) \(5k-2\): 3, 8, 13, 18, 23 ✓
(vi) \(5k-7\): \(k=2\): 3; \(k=3\): 8; \(k=4\): 13 ✓
Answer: (iii), (v), (vi) all capture the same sequence of numbers.
Similarly: consider \(5k-2\). What values does it take for different values of \(k\)? Numbers that leave a remainder of 3 when divided by 5 can also be seen as 2 less than multiples of 5, so \(5k-2\) with \(k \geq 1\).
\(k=1 \to 3\), \(k=2 \to 8\), \(k=3 \to 13\), \(k=4 \to 18\), \(k=5 \to 23\). All leave remainder 3 on division by 5. ✓

Figure it Out (Section 5.1)

Q1. The sum of four consecutive numbers is 34. Find these numbers.
Let the numbers be \(n, n+1, n+2, n+3\). Sum = \(4n+6 = 34 \Rightarrow n = 7\). Numbers: 7, 8, 9, 10.
Q2. Suppose \(p\) is the greatest of five consecutive numbers. Describe the other four numbers in terms of \(p\).
The five numbers: \(p-4, p-3, p-2, p-1, p\). Their sum = \(5p-10\).
Q3. For each statement, Always/Sometimes/Never True. Justify using algebra.
(i) Sum of two even numbers is a multiple of 3.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
(iii) The sum of two multiples of 6 and a multiple of 9 is a multiple of 3.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
(v) The sum of a multiple of 4 and a multiple of 12 is a multiple of 12.
(i) Sometimes. 4+8=12 ✓, but 2+4=6 is ÷3 but 4+6=10 is not. (ii) Sometimes. 9 is ÷9 but not ÷18. (iii) Always. \(6a + 6b + 9c = 3(2a+2b+3c)\). (iv) Always. \(6a+9b = 3(2a+3b)\). (v) Sometimes. 4+12=16 (not ÷12); 24+12=36 ✓.
Q4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Remainder 2 on ÷3 AND ÷4 means number minus 2 is divisible by both 3 and 4 → divisible by LCM(3,4)=12. Numbers: 2, 14, 26, 38, 50, ... Expression: \(12k + 2\).
Q5. "I hold some pebbles, two away. When I group them in 3's, one stays with me. Try pairing them up — it simply won't do. A rubber-band pebble remains in my view. Group them by 5, yet one's still around. But grouping by seven, perfection is found. More than one hundred would be far too bold. Can you tell me the number of pebbles I hold?"
The number \(N\) satisfies: \(N \equiv 1 \pmod 3\), \(N \equiv 1 \pmod 2\), \(N \equiv 1 \pmod 5\), \(N \equiv 0 \pmod 7\), and \(N \leq 100\). Common remainder 1 on ÷2, 3, 5 means \(N = 30k+1\): 1, 31, 61, 91. Check ÷7: 91 = 7 × 13 ✓. Answer: 91 pebbles.
Q6. Tathagat has written sentences that have the remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tathagat's claim true?
Each number has form \(6k+2\). Sum of three: \((6a+2)+(6b+2)+(6c+2) = 6(a+b+c) + 6 = 6(a+b+c+1)\). Yes, always a multiple of 6. Claim is true!
Q7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? (i) \(4779 - 661\)   (ii) \(4779 + 661\). Show the solution algebraically and visually.
\(661 = 7a+3\), \(4779 = 7b+5\). (i) \(4779-661 = 7(b-a) + 2\) → remainder 2. (ii) \(4779+661 = 7(a+b) + 8 = 7(a+b+1) + 1\) → remainder 1.
Q8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. Can you give a simple explanation of why it is the smallest?
Each condition says \(N+1\) is divisible by 3, 4, 5 respectively. So \(N+1\) is a multiple of LCM(3, 4, 5) = 60. Smallest: \(N = 59\). Check: 59 ÷ 3 = 19 R2 ✓, 59 ÷ 4 = 14 R3 ✓, 59 ÷ 5 = 11 R4 ✓.

Competency-Based Questions

Scenario: A school librarian arranges books on a shelf. When she places them in stacks of 4, 3 are left over. When she stacks them in 5, 4 are left over. When she stacks them in 6, 5 are left over. The school has between 50 and 100 books in this collection.
Q1. What is the total number of books?
L3 Apply
  • (a) 59
  • (b) 60
  • (c) 62
  • (d) 89
Answer: (a) 59. \(N+1\) divisible by 4, 5, 6, so \(N+1 = \text{LCM}(4,5,6) = 60\). Hence \(N = 59\).
Q2. Analyse why this trick works. If \(r_1, r_2, r_3\) are the remainders and \(d_1, d_2, d_3\) are the divisors, and if each remainder is exactly one less than its divisor, what general expression captures all such numbers?
L4 Analyse
Answer: When \(r_i = d_i - 1\) for every divisor, adding 1 to \(N\) makes it divisible by every \(d_i\). So \(N + 1\) is a multiple of LCM\((d_1, d_2, d_3)\). General expression: \(N = k \cdot \text{LCM}(d_1, d_2, d_3) - 1\).
Q3. Evaluate: Raj claims that 89 also satisfies this set of conditions. Verify or refute with reasoning.
L5 Evaluate
Answer: 89 ÷ 4 = 22 R1 (not R3). Claim false. The next valid number is \(2 \times 60 - 1 = 119\), which exceeds 100. So 59 is the unique solution in the given range.
Q4. Create a similar puzzle of your own: design a number-of-objects puzzle with 3 divisibility conditions and a range. Provide the answer and the reasoning.
L6 Create
Sample: "A crate holds mangoes. Packed in 5s: 2 left over. Packed in 7s: 2 left over. Packed in 9s: 2 left over. Fewer than 400. Find the number." \(N - 2\) divisible by LCM(5, 7, 9) = 315. So \(N = 317\). Many designs possible.

Assertion–Reason Questions

Assertion (A): The sum of four consecutive natural numbers is always even.
Reason (R): Two of every four consecutive numbers are odd and two are even, and odd + odd + even + even gives an even result.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Sum = \(4n+6\), always even. R explains A correctly.
Assertion (A): If a number is divisible by 12, it must be divisible by every factor of 12.
Reason (R): Divisibility is transitive: if \(a \mid b\) and \(c \mid a\), then \(c \mid b\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Divisibility is transitive. Every factor \(c\) of 12 divides any multiple of 12.
Assertion (A): Every even number can be written as 4p or 4p + 2 for some integer p.
Reason (R): Every integer when divided by 4 leaves a remainder of 0, 1, 2, or 3.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Even remainders are only 0 and 2, giving 4p and 4p+2. R explains A.

Frequently Asked Questions — Number Play

What is Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool in NCERT Class 8 Mathematics?

Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 5: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool step by step?

To solve problems on Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Number Play?

The essential formulas of Chapter 5 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool important for the Class 8 board exam?

Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool?

Common mistakes in Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 1 — Sums of Consecutive Numbers & Parity Patterns | Class 8 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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