Number Play
Number Play — Study Notes
NCERT-aligned · 10 notes · 3 shown free
Is This a Multiple Of?
ConceptIs This a Multiple Of?
This section introduces the exploration of sums of consecutive numbers and their properties. Anshu's observations about expressing numbers as sums of consecutive natural numbers spark several intriguing questions: Can every natural number be written as a sum of consecutive numbers? Which numbers can be expressed in more than one way as such sums? Are all odd numbers expressible as sums of two consecutive numbers? Can all even numbers be expressed as sums of consecutive numbers? Can zero be expressed as a sum of consecutive numbers, possibly involving negative numbers? These questions encourage students to investigate patterns and properties of numbers through experimentation and reasoning. The section further explores the parity (evenness or oddness) of expressions formed by placing '+' and '−' signs between four consecutive numbers. By systematically listing all eight possible expressions formed by four consecutive numbers and evaluating their sums, students observe that the results always have the same parity — specifically, they are always even numbers. This observation is generalized using algebraic reasoning, showing that switching signs changes the sum by an even number, thus preserving parity. The parity rules for sums and differences of odd and even numbers are revisited to support this conclusion. The section also hints at alternative explanations using positive and negative token models and encourages students to ponder whether this parity phenomenon holds for any set of four numbers, not just consecutive ones. This inquiry nurtures mathematical curiosity and introduces the power of algebraic reasoning to prove general properties without exhaustive computation.
- Sum of consecutive numbers can represent many natural numbers in multiple ways.
- Expressions formed by placing '+' and '−' signs between four consecutive numbers always have the same parity.
- Switching a '+' to '−' or vice versa changes the sum by an even number, preserving parity.
- Parity rules for sums and differences of odd and even numbers explain these observations.
- Algebraic reasoning generalizes the parity property for expressions involving four numbers.
- Mathematical reasoning can prove properties without checking each case individually.
- 📌 Parity: The property of an integer being even or odd.
- 📌 Consecutive numbers: Numbers that follow each other in order without gaps.
- 📌 Algebraic reasoning: Using algebraic expressions and operations to generalize mathematical properties.
Breaking Even
ConceptBreaking Even
This section focuses on identifying even numbers through arithmetic expressions and algebraic forms. It begins by asking students to determine which given arithmetic expressions evaluate to even numbers without performing full calculations. This exercise leverages the understanding of parity and arithmetic operations. The section then extends this understanding to algebraic expressions involving variables. It explains that expressions like 4m + 2q are always even for any integers m and q because both terms are multiples of 2, and their sum is also a multiple of 2. This is justified both by parity reasoning and by factoring out 2: 4m + 2q = 2(2m + q). The section also discusses expressions that do not always yield even numbers, such as x² + 2, where the parity depends on whether x is even or odd. Examples illustrate when such expressions are even or odd. Students are encouraged to analyze other algebraic expressions to determine if they always produce even numbers, providing examples and counterexamples. They are also invited to create their own algebraic expressions that always evaluate to even numbers. This section strengthens students' ability to reason about parity in algebraic contexts, preparing them for deeper exploration of multiples, divisibility, and number properties.
- Arithmetic expressions can be analyzed for parity without full calculation.
- Expressions like 4m + 2q are always even because they factor as 2(2m + q).
- Not all algebraic expressions always yield even numbers; parity depends on variable values.
- Examples and counterexamples help understand when expressions are always even.
- Factoring and parity rules are key tools in analyzing algebraic expressions.
- Students are encouraged to create algebraic expressions that always produce even numbers.
- 📌 Even number: An integer divisible by 2.
- 📌 Parity: The classification of integers as even or odd.
- 📌 Algebraic expression: A mathematical phrase involving variables and constants.
Pairs to Make Fours
ConceptPairs to Make Fours
This section investigates the conditions under which the sum of two even numbers is divisible by 4. Students experiment by adding pairs of even numbers and observing whether their sum is a multiple of 4. Through exploration, they discover that even n
Practice Questions — Number Play
Includes NCERT exercise questions with answers
Q1.6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Answer:
3p7q8 is divisible by 44, means divisible by 11 and 4. For divisibility by 4, last two digits must be divisible by 4. So possible q8 are 08, 28, 48, 68, 88. For divisibility by 11, difference between sum of the odd place digits and even place digits must be 0 or multiple of 11. Sum of odd place digits = 8 + 7 + 3 = 18. Sum of even place digits = p + q. Difference is 18 − (p + q). Let k = 18 − (p + q) = 0 or a multiple of 11. (i) if p + q = 18 Not possible for q (0, 2, 4, 6, 8), since p is a digit. (ii) if 18 – (p + q) = 11 then p + q = 7. Possible pairs: (p=7, q=0), (p=5, q=2), (p=3, q=4), (p=1, q=6). (iii) 18 – (p + q) cannot be other multiples of 11. Hence the possible pairs for p and q are (7,0), (5,2), (3,4), and (1,6).
Explanation:
Step 1: Divisibility by 4 requires last two digits q8 to be divisible by 4, so q8 can be 08, 28, 48, 68, 88. Step 2: Divisibility by 11 requires difference between sum of digits in odd and even positions to be 0 or multiple of 11. Sum odd = 3 + 7 + 8 = 18. Sum even = p + q. Difference = 18 - (p + q). Check for difference = 0 or ±11, ±22,... Only difference = 11 is possible with p + q = 7. Check all q from possible q values and find p accordingly. Thus, pairs are (7,0), (5,2), (3,4), (1,6).
Q2.7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer:
Let the three consecutive numbers be n, n+1, n+2. Given: n is multiple of 2, n+1 is multiple of 3, n+2 is multiple of 4. One such set is 2, 3, 4. Since the numbers are consecutive, the pattern repeats every LCM of 2, 3, and 4, which is 12. So, the next such set is 14, 15, 16. Hence, such numbers occur every 12 numbers.
Explanation:
Step 1: Assign variables to consecutive numbers. Step 2: Apply divisibility conditions. Step 3: Find one example: 2 (even), 3 (multiple of 3), 4 (multiple of 4). Step 4: The pattern repeats every LCM(2,3,4) = 12. Step 5: Next set is 14,15,16 and so on.
Q3.8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Answer:
Since 36 = 4 × 9, a number divisible by 36 must be divisible by both 4 and 9. For divisibility by 4, last two digits must be divisible by 4. For divisibility by 9, sum of digits must be divisible by 9. Between 45000 and 47000, multiples of 36 are: 45036, 45072, 45108, 45144, 45180. Approach: Check numbers ending with last two digits divisible by 4 and sum of digits divisible by 9.
Explanation:
Step 1: Understand divisibility rules for 4 and 9. Step 2: Check numbers in given range ending with digits divisible by 4. Step 3: Check sum of digits for divisibility by 9. Step 4: List first five such numbers.
Q4.9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Answer:
Let the five consecutive even numbers be: 5p - 4, 5p - 2, 5p, 5p + 2, 5p + 4. Since the middle number is 5p, the two numbers before it are 2 and 4 less, and the two numbers after it are 2 and 4 more, respectively.
Explanation:
Step 1: Consecutive even numbers differ by 2. Step 2: Middle number is 5p. Step 3: Numbers before middle are 5p - 2 and 5p - 4. Step 4: Numbers after middle are 5p + 2 and 5p + 4.
Q5.10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Answer:
A number divisible by 15 must be divisible by 3 and 5. So, it must end with 0 or 5. If it ends with 0, reversed number starts with 0, which is not a 6-digit number. So, last digit f = 5. For reversed number to be divisible by 6, it must be divisible by 2 and 3. So, first digit a of original number (which is last digit of reversed) must be even and non-zero. Possible a values: 2, 4, 6, 8. Examples: 200025, 200055, 200085, 202005, etc., where sum of digits is divisible by 3. Hence, such numbers exist with these conditions.
Explanation:
Step 1: Divisibility by 15 requires divisibility by 3 and 5. Step 2: Number ends with 0 or 5. Step 3: If ends with 0, reversed number not 6-digit. Step 4: So ends with 5. Step 5: Reversed number divisible by 6 requires even first digit. Step 6: Choose a = 2,4,6,8. Step 7: Check sum of digits divisible by 3. Step 8: Construct numbers accordingly.
Q6.11. Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.
Answer:
Deepak’s conjecture is false. Let n be a multiple of 11, so n = 11k. When doubled, 2n = 2 × 11k = 11 × (2k), which is also a multiple of 11. Hence, all multiples of 11 when doubled remain multiples of 11.
Explanation:
Step 1: Represent multiple of 11 as 11k. Step 2: Double it: 2 × 11k = 11 × 2k. Step 3: Since 2k is an integer, 2n is multiple of 11. Step 4: Therefore, Deepak’s claim that some multiples of 11 when doubled are not multiples of 11 is incorrect.
Q7.12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers will be divisible by 6. (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. (iv) 8 (7b – 3) – 4 (11b + 1) is a multiple of 12.
Answer:
(i) Always true. (ii) Always true. (iii) Always true. (iv) Never true.
Explanation:
(i) Product of multiples of 6 and 3 includes at least two 3s and one 2, so multiple of 9. (ii) Sum of three consecutive even numbers is divisible by 6 because sum is 6 times the middle number. (iii) Rearranging digits does not affect divisibility by 6 if digits sum and last digit conditions hold. (iv) Expression simplifies to a form not divisible by 12 for all b, so never true.
Q8.Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer:
Let the three consecutive numbers be n, n + 1, n + 2. Given that n is a multiple of 2, n + 1 is a multiple of 3, and n + 2 is a multiple of 4. One such set is 2, 3, 4. Since the numbers are consecutive, the pattern repeats every LCM of 2, 3, and 4, which is 12. Therefore, the next such set is 14, 15, 16, and so on, repeating every 12 numbers.
Explanation:
We check the conditions for divisibility for consecutive numbers and find the pattern repeats every 12 numbers because 12 is the LCM of 2, 3, and 4.
All 7 Chapters in Ganita Prakash Part-I
Mathematics · Class 8