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**Definition:** Consecutive numbers are natural numbers that follow one another in order (e.g., 3, 4, 5, 6 or 7, 8, 9).
**Key Question:** Can every natural number be written as a sum of consecutive numbers? The answer is NO — not every number can be expressed this way.
**Examples of numbers as sums of consecutive numbers:**
**Important Observation:** Some numbers can be written as sums of consecutive numbers in multiple ways (like 15), while others cannot be written this way at all (like powers of 2: 2, 4, 8, 16...).
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**Activity:** Take any 4 consecutive numbers and place '+' and '–' signs in between them.
For example, with 3, 4, 5, 6:
**Total number of sign combinations:** For 4 consecutive numbers, there are 8 different ways to place '+' and '–' signs (2³ = 8 combinations).
**Amazing Discovery: All Results Are Even**
When 4 consecutive numbers are chosen, NO MATTER HOW the '+' and '–' signs are placed between them, the resulting expressions ALWAYS produce EVEN numbers (including negative even numbers like –2, –4, –6, –12, etc.).
**This pattern holds for ANY set of 4 consecutive numbers.**
Example with 5, 6, 7, 8:
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**Theorem:** When one sign in an expression changes (from + to – or vice versa), the value of the expression changes by an EVEN number.
**Proof:** Consider the expression: a + b – c – d
If we replace +b with –b, we get: a – b – c – d
The change in value is:
(a + b – c – d) – (a – b – c – d)
= a + b – c – d – a + b + c + d
= 2b (which is EVEN)
**Key Deduction:** If two numbers differ by an even number, they must have the SAME PARITY (both even or both odd).
**Application:** Starting from ANY expression using 4 numbers and signs, we can reach all 8 possible expressions by switching signs one at a time. Since each switch changes the result by an even number, ALL 8 expressions have the SAME PARITY.
Since at least one expression (e.g., a + b + c + d) is even, ALL expressions are even.
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**Fundamental Parity Rules:**
**Key Insight:** a + b and a – b ALWAYS have the SAME parity, regardless of whether a and b are even or odd.
**Proof of the Key Insight:**
**Generalization:** By extension, a ± b ± c ± d ALWAYS has the SAME PARITY for any choice of signs.
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The positive and negative token model (from the Integers chapter) can be used to visualize this property:
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**Question:** Is the "all same parity" phenomenon limited to 4 numbers?
**Answer:** No, it generalizes to ANY even number of consecutive integers.
**For odd number of terms (3, 5, 7, ...):** The parities can vary.
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**Task:** Determine which algebraic expressions ALWAYS give an even number for ANY integer values of the variables.
An algebraic expression always produces even numbers if it can be written as 2 × (something).
**Examples that ALWAYS give even:**
**Examples that DON'T always give even:**
Determine if each term in the expression is always even or always odd, then apply parity rules.
**Example:** 4m + 2n
**Example:** x² + 2
**Test divisibility or compute if necessary:**
**Always Even:**
**Note:** When checking arithmetic expressions, identify the parity of each number first:
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**Observation:** Not all even numbers are multiples of 4.
**Classification of Even Numbers:**
**Why is the remainder always 2 for even non-multiples of 4?**
Even numbers not divisible by 4 can be written as 4q + 2 (where q is any non-negative integer).
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**Expression:** 4p + 4q (where p and q are integers)
**Algebraic Form:** 4p + 4q = 4(p + q)
**Result:** ALWAYS a multiple of 4 ✓
**Examples:**
**Visual Explanation:** If we represent 4p as p rows of 4 items and 4q as q rows of 4 items, their sum is (p+q) rows of 4 items, which is clearly a multiple of 4.
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**Expression:** (4p + 2) + (4q + 2) (where p and q are integers)
**Algebraic Expansion:**
(4p + 2) + (4q + 2)
= 4p + 4q + 4
= 4(p + q + 1)
**Result:** ALWAYS a multiple of 4 ✓
**Why This Works:** The two remainders of 2 combine to form an additional group of 4.
**Examples:**
**Visual Explanation:** Two "half-groups" (each 2 short of a complete group of 4) combine to form one complete group of 4. So the total is (p+q+1) groups of 4.
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**Expression:** 4p + (4q + 2) (where p and q are integers)
**Algebraic Expansion:**
4p + (4q + 2)
= 4p + 4q + 2
= 4(p + q) + 2
**Result:** ALWAYS leaves remainder 2 when divided by 4 (NOT a multiple of 4) ✗
**Why:** A complete group of 4 combined with a "half-group" of 2 still leaves a remainder of 2.
**Examples:**
**Visual Explanation:** The complete rows plus the partial row still leave one partial group.
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**Pattern:** Sum is a multiple of 4 if and only if BOTH numbers are either:
1. Both multiples of 4, OR
2. Both non-multiples of 4 (both leaving remainder 2)
**This is analogous to parity rules:** Just as odd + odd = even, here "remainder-2" + "remainder-2" = "remainder-0" (multiple of 4).
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**Three Categories:**
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**Claim:** ALWAYS TRUE ✓
**Algebraic Proof:** If 8|M and 8|N, then:
Since M + N = 8(a + b), it is divisible by 8.
**Does this hold for subtraction?** YES, ALWAYS TRUE ✓
**Examples:**
**General Theorem:** If a divides M and a divides N, then a divides both (M + N) and (M – N).
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**Claim:** SOMETIMES TRUE ✓
**Algebraic Analysis:** 8m (a multiple of 8) can be expressed as:
**Examples:**
**Conclusion:** The statement is only sometimes true because a number divisible by 8 can be decomposed in multiple ways.
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**Claim:** ALWAYS TRUE ✓
**Algebraic Proof:** If 7|j, then j = 7k for some integer k.
Any multiple of j is: j × m = (7k) × m = 7(km)
Since this equals 7 times an integer (km), it is divisible by 7.
**Examples:**
**General Theorem:** If A is divisible by k, then all multiples of A are divisible by k.
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**Claim:** ALWAYS TRUE ✓
**Factors of 12:** 1, 2, 3, 4, 6, 12
**Algebraic Proof:** If 12|M, then M = 12m for some integer m.
Since 12 = 2² × 3, any multiple of 12 contains these prime factors.
For any factor f of 12:
where q is such that f × q = 12.
Thus M is divisible by every factor of 12.
**Examples:**
**General Theorem:** If A is divisible by k, then A is divisible by all factors of k.
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**Claim:** SOMETIMES TRUE ✓
**Algebraic Analysis:** If 7|M, then M = 7k for some integer k.
M is divisible by 7m (a multiple of 7) if and only if:
M ÷ 7m = (7k) ÷ 7m = k/m is an integer
This means m must be a FACTOR of k.
**Examples:**
**Conclusion:** Sometimes true, depending on whether the divisor's factor divides the quotient k.
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**Claim:** ALWAYS TRUE ✓
**Reasoning:** 9 and 4 are coprime (their GCD is 1).
If a number is divisible by both 9 and 4, it must be divisible by 9 × 4 = 36.
**Algebraic Proof:** If 9|M and 4|M, then:
Thus 36|M.
**Examples:**
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**Claim:** SOMETIMES TRUE ✓
**Reasoning:** 6 and 4 are NOT coprime (GCD = 2).
**LCM Method:** LCM(6, 4) = 12, not 24.
**Examples:**
**General Rule:** If A is divisible by both k and m, then A is divisible by LCM(k, m), not necessarily by k × m.
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**Claim:** NEVER TRUE ✗
**Reasoning 1 (Parity Argument):**
Therefore, this is NEVER true.
**Reasoning 2 (Algebraic Proof):** Suppose (2n) + (2m + 1) = 6j
Then: 2n + 2m + 1 = 6j
2(n + m) + 1 = 6j
2(n + m) = 6j – 1
The left side is EVEN, but the right side (6j – 1) is ODD (since 6j is even).
An even number can never equal an odd number, so this equation has no solution.
**Examples attempting to verify:**
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**Question:** Find a number that leaves a remainder of 3 when divided by 5. Write an algebraic expression for ALL such numbers.
**Numbers leaving remainder 3 when divided by 5:**
**General Form:** 5k + 3 (where k = 0, 1, 2, 3, ...)
**Verification:** These numbers are 3 MORE than multiples of 5.
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**Can the same set be written differently?**
Consider 5k – 2 where k ≥ 1:
**Why This Works:** 5k – 2 is the same as 5(k – 1) + 5 – 2 = 5(k – 1) + 3
So 5k – 2 (with k ≥ 1) represents the SAME set as 5k + 3 (with k ≥ 0).
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**Given options:** (i) 3k + 5, (ii) 3k – 5, (iii) 3k/5, (iv) 5k + 3, (v) 5k – 2, (vi) 5k – 3
**Correct answers:** (iv) 5k + 3 and (v) 5k – 2 (both with appropriate domain restrictions)
**Why others don't work:**
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**Let the four consecutive numbers be:** n, n+1, n+2, n+3
**Set up the equation:**
n + (n + 1) + (n + 2) + (n + 3) = 34
**Simplify:**
4n + 6 = 34
4n = 28
n = 7
**The four numbers are:** 7, 8, 9, 10
**Verification:** 7 + 8 + 9 + 10 = 34 ✓
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**Let the greatest number be:** p
Then the other four consecutive numbers, in decreasing order, are:
**Or listing from smallest to largest:**
p – 4, p – 3, p – 2, p – 1, p
**Verification:** These are indeed 5 consecutive numbers with p as the greatest.
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#### (i) The Sum of Two Even Numbers Is a Multiple of 3
**Claim:** SOMETIMES TRUE ✓
**Explanation:** Even numbers are 2k for integer k. The sum of two even numbers is 2m + 2n = 2(m + n).
Whether this is a multiple of 3 depends on whether (m + n) is a multiple of 3.
**Examples:**
**Conclusion:** Sometimes true, sometimes false.
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#### (ii) If a Number Is Not Divisible by 18, Then It's Not Divisible by 9
**Claim:** NEVER TRUE (This is a false logical statement) ✗
**Counterexample:** 9 itself
**Another example:** 27
**Logical Error:** The contrapositive would be: "If not divisible by 9, then not divisible by 18," which IS true. But the original statement reverses this.
**Algebraic Explanation:** If a number is divisible by 9, it's of the form 9k. For it to be divisible by 18, we need 9k to be divisible by 18 = 9 × 2, which means k must be even. But k could be odd.
**Conclusion:** Never true (or more precisely, the logical implication is false).
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#### (iii) If Two Numbers Are Not Divisible by 6, Their Sum Is Not Divisible by 6
**Claim:** SOMETIMES TRUE ✓
**Explanation:** If two numbers are not individually divisible by 6, their sum could still be divisible by 6.
**Examples:**
**Algebraic Analysis:** Let two numbers not divisible by 6 be of the form 6p + r₁ and 6q + r₂, where r₁, r₂ ≠ 0.
Their sum: (6p + r₁) + (6q + r₂) = 6(p + q) + (r₁ + r₂)
This sum is divisible by 6 if and only if r₁ + r₂ is divisible by 6.
**Conclusion:** Sometimes true (when r₁ + r₂ is divisible by 6), sometimes false.
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#### (iv) The Sum of a Multiple of 6 and a Multiple of 9 Is a Multiple of 3
**Claim:** ALWAYS TRUE ✓
**Algebraic Proof:**
Since the sum = 3 × (integer), it's divisible by 3.
**Explanation:** Both 6 and 9 are multiples of 3, so their sum must also be a multiple of 3.
**Examples:**
**Conclusion:** Always true.
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#### (v) The Sum of a Multiple of 6 and a Multiple of 3 Is a Multiple of 9
**Claim:** SOMETIMES TRUE ✓
**Explanation:** Not all multiples of 3 when added to multiples of 6 give multiples of 9.
**Algebraic Analysis:**
For this to be divisible by 9 = 3², we need (2a + b) to be divisible by 3.
**Examples:**
Q1. When 4 consecutive numbers are chosen and combined with '+' and '–' signs in any way, what can we always say about the resulting expressions?
Answer: A — According to the study material, when four consecutive numbers are combined with any combination of '+' and '–' signs, all resulting expressions always have the same parity because switching a sign changes the value by an even number.
Q2. Which of the following statements about even numbers is correct?
Answer: B — The study material clearly states that even numbers are of two types: multiples of 4 (remainder 0) and non-multiples of 4 (remainder 2).
Q3. Which of the following is NOT correct about parity under operations?
Answer: C — According to the parity rules in the study material, odd ± even = odd, not even. Options A, B, and D are all correct parity rules.
Q4. Consider the expression 4m + 2n where m and n are any integers. Which statement is true?
Answer: B — The expression 4m + 2n can be factored as 2(2m + n), which means 2 is always a factor, guaranteeing the result is always even regardless of integer values of m and n.
Q5. If we take the expression a + b – c – d and replace +b with –b to get a – b – c – d, by how much has the value changed?
Answer: A — The difference between (a + b – c – d) and (a – b – c – d) is 2b, which is always even because it has 2 as a factor.
Q6. Rajesh is checking which arithmetic expressions are even without calculating them. Which of these expressions is definitely even?
Answer: B — Using parity rules: odd + odd = even. Option A: even – even = even ✓, but option B is also correct. However, 43 + 37 = 80 (even), confirming odd + odd = even.
Q7. Which of the following algebraic expressions will NOT always give an even number?
Answer: D — 6m – 3n = 3(2m – n). Since 3 is an odd factor, if (2m – n) is odd, the result is odd. For example, if m = 1 and n = 0, then 6(1) – 3(0) = 6 (even), but if m = 0 and n = 1, then 6(0) – 3(1) = –3 (odd).
Q8. In a market, shopkeeper Meena has two piles of coins. One pile has coins that are multiples of 4 (like 4, 8, 12...) and another pile has coins that leave remainder 2 when divided by 4 (like 2, 6, 10...). If she picks one coin from each pile and adds them, what can we conclude?
Answer: C — When a multiple of 4 (remainder 0) is added to a number with remainder 2, the sum will have remainder 2, similar to how even + odd = odd in parity.
Q9. The expression x² + 2 gives an even number in some cases and an odd number in others. When does it give an odd number?
Answer: C — If x is odd, then x² is odd. Odd + even (2) = odd. For example, if x = 3, then x² + 2 = 9 + 2 = 11 (odd). If x is even, x² is even, so even + 2 = even.
Q10. Anshu wrote: 'All odd numbers can be written as the sum of two consecutive numbers.' Is this statement correct, and why?
Answer: A — Any two consecutive numbers are of the form n and (n+1), where one is even and one is odd. Since even + odd = odd, the sum is always odd, confirming that any odd number can be written this way (for example, 7 = 3 + 4).
What is the parity of a + b when a is even and b is even?
The parity is always even, regardless of the specific values of a and b.
Which two types of even numbers can be classified based on division by 4?
Multiples of 4 (remainder 0) and non-multiples of 4 (remainder 2).
If 8 divides two numbers M and N separately, will 8 always divide M + N?
Yes, because 8a + 8b = 8(a + b), so the sum is always a multiple of 8.
What happens to the parity of an expression when you switch one '+' sign to a '−' sign among four consecutive numbers?
The parity remains the same because the expression changes by an even number (2b).
Can every odd number be written as a sum of two consecutive numbers?
Yes, because consecutive numbers n and (n+1) sum to 2n+1, which is always odd.
Why is 4m + 2q always even for any integers m and q?
Because 4m + 2q = 2(2m + q), which has 2 as a factor, making it always even.
If a number is divisible by 12, will it always be divisible by 3?
Yes, because 3 is a factor of 12, so all multiples of 12 are divisible by 3.
When will the sum of two even numbers give a multiple of 4?
When both are multiples of 4, or when both are non-multiples of 4 (remainder 2 each).
What is the algebraic form of any number divisible by 8?
The form is 8a, where a is any integer.
If a divides M and a divides N, what can you say about a and M − N?
The number a also divides M − N because M − N = a(b − c) for some integers b and c.
Write the algebraic form of any number divisible by 7. [1 mark]
Use a variable to represent any integer multiple; the form is 7a where a is any integer.
Explain why the sum of two numbers that are both multiples of 6 is always a multiple of 6. Give one numerical example. [2 marks]
Express both numbers as 6m and 6n; add them to get 6(m + n); example: 12 + 18 = 30 = 6 × 5.
Take four consecutive numbers: 5, 6, 7, 8. Write three different expressions using + and − signs. Show that all three expressions have the same parity (all even or all odd). [3 marks]
Examples: 5 + 6 + 7 + 8 = 26 (even), 5 + 6 − 7 − 8 = −4 (even), 5 − 6 − 7 + 8 = 0 (even); use parity rules to explain that switching signs changes the value by an even amount.
State whether the following statement is always true, sometimes true, or never true. Justify your answer with algebraic reasoning and provide at least one example and one counterexample where applicable: 'If a number is divisible by 8, then it is also divisible by any multiple of 8.' [5 marks]
This statement is sometimes true; a number divisible by 8 (like 8m) is divisible by 16 only if m is even (example: 24 = 8 × 3, not divisible by 16; counterexample: 32 = 8 × 4, is divisible by 16); use algebraic form to show when and when not.
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