StudyOSStudyOS CBSE Study Material, Class 5–12

Number Play

NCERT Class 8 · Mathematics Based on NCERT Class 8 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explores patterns and properties of numbers, focusing on sums of consecutive numbers, parity (even and odd), divisibility rules, digital roots, and cryptarithms. The big idea is that numbers follow predictable patterns based on their factors, remainders, and digit sums, which can be discovered through reasoning and algebra. Students will learn to determine divisibility quickly, understand why these shortcuts work, and solve puzzles involving hidden digits.
Key Concepts (22)
Consecutive numbers
Numbers that follow each other in order, each increasing by 1 from the previous number.
e.g. 3, 4, 5, 6 are four consecutive numbers.
Parity
Whether a number is even or odd. Even numbers are divisible by 2; odd numbers are not.
e.g. 4 is even, 7 is odd.
Even number
A number that is divisible by 2, leaving no remainder. It can be written as 2 × k for some integer k.
e.g. 12, -4, 0 are even numbers.
Odd number
A number that is not divisible by 2, leaving a remainder of 1 when divided by 2. It can be written as 2k + 1.
e.g. 7, -3, 15 are odd numbers.
Multiple
A number that can be divided by another number without leaving a remainder. For example, 15 is a multiple of 3 and 5.
e.g. 24 is a multiple of 8 because 24 ÷ 8 = 3 exactly.
Divisibility
A number is divisible by another if the division leaves no remainder.
e.g. 36 is divisible by 9 because 36 ÷ 9 = 4 exactly.
Divisibility rule for 2
A number is divisible by 2 if its units digit is 0, 2, 4, 6, or 8 (i.e., it is even).
e.g. 128 is divisible by 2 because its units digit is 8.
Divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
e.g. 123: 1+2+3 = 6, which is divisible by 3, so 123 is divisible by 3.
Divisibility rule for 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
e.g. 128: last two digits are 28, and 28 ÷ 4 = 7, so 128 is divisible by 4.
Divisibility rule for 5
A number is divisible by 5 if its units digit is 0 or 5.
e.g. 990 is divisible by 5 because its units digit is 0.
Divisibility rule for 6
A number is divisible by 6 if it is divisible by both 2 and 3.
e.g. 186 is divisible by 2 (units digit 6) and by 3 (1+8+6=15, divisible by 3), so it is divisible by 6.
Divisibility rule for 8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
e.g. 128: last three digits are 128, and 128 ÷ 8 = 16, so 128 is divisible by 8.
Divisibility rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
e.g. 405: 4+0+5 = 9, which is divisible by 9, so 405 is divisible by 9.
Divisibility rule for 10
A number is divisible by 10 if its units digit is 0.
e.g. 990 is divisible by 10 because its units digit is 0.
Divisibility rule for 11
A number is divisible by 11 if the difference between the sum of digits in odd places (from the right) and the sum of digits in even places is 0 or a multiple of 11.
e.g. For 462: (2+4) - 6 = 0, so 462 is divisible by 11.
Digital root
The single-digit number obtained by repeatedly adding the digits of a number until a single digit is obtained.
e.g. For 489710: 4+8+9+7+1+0 = 29, 2+9 = 11, 1+1 = 2. The digital root is 2.
Cryptarithm
A puzzle where each letter stands for a distinct digit, each digit is represented by at most one letter, and the first digit of a number is never 0.
e.g. A1 + 1B = B0. Here A and B are digits to be found.
Remainder
The amount left over after division when one number does not divide the other exactly.
e.g. When 7 is divided by 3, the quotient is 2 and the remainder is 1.
Factor
A number that divides another number exactly, leaving no remainder.
e.g. 1, 2, 3, 4, 6, and 12 are factors of 12.
Multiple of a number
The product of that number with any integer.
e.g. Multiples of 7 include 7, 14, 21, 28, ...
LCM (Least Common Multiple)
The smallest positive integer that is a multiple of two or more numbers.
e.g. LCM of 6 and 4 is 12.
Prime factorisation
Expressing a number as a product of its prime factors.
e.g. 24 = 2 × 2 × 2 × 3 = 2³ × 3.
Formulas & Equations (10)
Even number = 2k
Reads as: An even number equals 2 times k
Variables: k is any integer
Use when: Representing any even number algebraically
⚠ Students forget that k can be negative or zero, not just positive.
Odd number = 2k + 1
Reads as: An odd number equals 2 times k plus 1
Variables: k is any integer
Use when: Representing any odd number algebraically
⚠ Students write 2k - 1 and forget that for k=0, 2(0)-1 = -1 is also odd, but the standard form is 2k+1.
Number = ... + 1000d + 100c + 10b + a
Reads as: A number equals the sum of its digits multiplied by their place values
Variables: a is units digit, b is tens digit, c is hundreds digit, d is thousands digit, etc.
Use when: Explaining divisibility rules using algebra
⚠ Students forget to include the highest place value terms and write only up to hundreds.
Divisibility by 9: Sum of digits divisible by 9
Reads as: A number is divisible by 9 if and only if the sum of its digits is divisible by 9
Variables: No variables; it is a rule
Use when: Checking if a number is divisible by 9 without dividing
⚠ Students think the rule works for divisibility by 3 in the same way but forget that the sum must be divisible by 3, not 9.
Divisibility by 11: (Sum of digits in odd places) - (Sum of digits in even places) = 0 or multiple of 11
Reads as: The difference between the sum of digits in odd positions and even positions must be 0 or a multiple of 11
Variables: Odd places are units, hundreds, ten-thousands, etc.; even places are tens, thousands, etc.
Use when: Checking divisibility by 11
⚠ Students start counting from the left instead of the right (units place).
Number leaving remainder r when divided by d = d × k + r
Reads as: A number that leaves remainder r when divided by d can be written as d times k plus r
Variables: d is divisor, k is any integer (quotient), r is remainder (0 ≤ r < d)
Use when: Describing numbers with a given remainder
⚠ Students forget that remainder must be less than the divisor.
If a divides M and a divides N, then a divides M + N and a divides M - N
Reads as: If a is a factor of both M and N, then a is also a factor of their sum and difference
Variables: a, M, N are integers
Use when: Proving properties of multiples and factors
⚠ Students think this works for any M and N, but it only works if a divides both M and N individually.
If A is divisible by k, then all multiples of A are divisible by k
Reads as: If A is a multiple of k, then any multiple of A is also a multiple of k
Variables: A and k are integers
Use when: Reasoning about divisibility of products
⚠ Students think the converse is also true: if a multiple of A is divisible by k, then A is divisible by k. This is false.
If A is divisible by k, then A is divisible by all factors of k
Reads as: If A is a multiple of k, then A is also a multiple of every factor of k
Variables: A and k are integers
Use when: Checking divisibility by factors of a known divisor
⚠ Students think the reverse: if A is divisible by a factor of k, then A is divisible by k. This is false (e.g., 12 is divisible by 4 but not by 8).
If A is divisible by k and A is divisible by m, then A is divisible by LCM(k, m)
Reads as: If A is a multiple of both k and m, then A is a multiple of their least common multiple
Variables: A, k, m are integers
Use when: Determining the smallest number divisible by two given numbers
⚠ Students think A must be divisible by the product k × m, which is not always true (e.g., 12 is divisible by 4 and 6, but not by 24).
How It Works
Finding sum of consecutive numbers
  1. Identify the starting number and how many consecutive numbers to add.
  2. Add the numbers in order: first + second + third + ...
  3. Alternatively, use the formula: sum = (first + last) × (number of terms) ÷ 2.
💡 Odd numbers can always be written as sum of two consecutive numbers (e.g., 7 = 3 + 4). Not all even numbers can be written as sum of consecutive numbers.
Determining parity of expressions with + and - signs on four consecutive numbers
  1. Take any four consecutive numbers, say a, a+1, a+2, a+3.
  2. Place + and - signs in all eight possible combinations between them.
  3. Evaluate each expression.
  4. Observe that all results are even numbers (parity is same for all eight expressions).
💡 Switching any sign changes the value by an even number, so all eight expressions have the same parity (all even or all odd). For four consecutive numbers, they are always even.
Checking divisibility by 9 using digit sum
  1. Add all the digits of the number.
  2. If the sum is divisible by 9, the original number is divisible by 9.
  3. If the sum is not divisible by 9, the remainder when the sum is divided by 9 is the same as the remainder of the original number divided by 9.
  4. Repeat adding digits of the sum until a single digit is obtained (digital root). If the digital root is 9, the number is divisible by 9.
💡 This works because every place value (10, 100, 1000, ...) is 1 more than a multiple of 9.
Checking divisibility by 11 using alternating sum
  1. Write the number and label the digits from the rightmost (units) as position 1, tens as position 2, hundreds as position 3, etc.
  2. Add the digits in odd positions (1st, 3rd, 5th, ...).
  3. Add the digits in even positions (2nd, 4th, 6th, ...).
  4. Subtract the sum of even positions from the sum of odd positions.
  5. If the result is 0 or a multiple of 11, the number is divisible by 11.
💡 Always start counting from the units place (rightmost digit) as position 1.
Finding digital root of a number
  1. Add all the digits of the number.
  2. If the sum is a single digit, that is the digital root.
  3. If the sum has more than one digit, add its digits again.
  4. Repeat until a single digit is obtained.
💡 The digital root of a multiple of 9 is always 9 (except for 0 itself).
Solving a cryptarithm
  1. Identify that each letter stands for a unique digit (0-9).
  2. The first digit of any number cannot be 0.
  3. Use the given operation (addition, multiplication) to set up equations.
  4. Use properties like parity, divisibility, and digit constraints to narrow down possibilities.
  5. Test possible digits systematically until all conditions are satisfied.
💡 Always check that the first digit of each number is not 0, and that no two letters represent the same digit.
Checking divisibility by 24 using factors 3 and 8
  1. Check if the number is divisible by 3 (sum of digits divisible by 3).
  2. Check if the number is divisible by 8 (last three digits divisible by 8).
  3. If both conditions are true, the number is divisible by 24.
  4. This works because 3 and 8 are coprime (their LCM is 24).
💡 Checking divisibility by 4 and 6 does not guarantee divisibility by 24 because 4 and 6 are not coprime (LCM of 4 and 6 is 12, not 24).
Exam Traps — Don't Make These Mistakes
⚠ Confusing divisibility by 3 and 9
✗ If the sum of digits is divisible by 3, the number is divisible by 9.
→ Divisibility by 3 requires sum divisible by 3; divisibility by 9 requires sum divisible by 9. A number like 15 has digit sum 6 (divisible by 3) but is not divisible by 9.
✓ A number is divisible by 9 only if the sum of its digits is divisible by 9. If the sum is divisible by 3 but not 9, the number is divisible by 3 but not by 9.
⚠ Incorrect counting for divisibility by 11
✗ Add digits from left to right, alternating signs starting with +.
→ The rule requires starting from the units place (rightmost digit) as position 1. Starting from the left gives a different result for numbers with an even number of digits.
✓ Label digits from the rightmost (units) as position 1. Add digits in odd positions, subtract digits in even positions. The result must be 0 or a multiple of 11.
⚠ Assuming sum of two even numbers is always a multiple of 4
✗ The sum of any two even numbers is always divisible by 4.
→ Even numbers can be of two types: multiples of 4 (like 4, 8) and even numbers not multiples of 4 (like 2, 6). Sum of two multiples of 4 is a multiple of 4. Sum of two even numbers not multiples of 4 is also a multiple of 4. But sum of a multiple of 4 and an even number not multiple of 4 is not a multiple of 4 (e.g., 4 + 2 = 6).
✓ The sum of two even numbers is always even, but it is a multiple of 4 only if both numbers are multiples of 4, or both are even but not multiples of 4. If one is a multiple of 4 and the other is not, the sum is not a multiple of 4.
⚠ Misapplying 'if a divides M and N, then a divides M+N'
✗ If 8 divides 16 and 8 divides 24, then 8 divides 16 + 24 = 40. This is always true for any numbers.
→ The statement is true only if a divides both M and N individually. Students often apply it when a divides only one of the numbers, which is incorrect.
✓ The property is: If a divides M and a divides N, then a divides M + N and a divides M - N. Both M and N must be multiples of a for the sum/difference to be a multiple of a.
⚠ Confusing 'multiple of' and 'factor of'
✗ If a number is divisible by 12, it is divisible by all multiples of 12.
→ If a number is divisible by 12, it is divisible by all factors of 12 (like 2, 3, 4, 6), not by all multiples of 12 (like 24, 36). For example, 12 is divisible by 12 but not by 24.
✓ If a number is divisible by k, then it is divisible by all factors of k, but not necessarily by multiples of k.
Compare & Contrast
Condition for divisibility
Divisibility by 3
Divisibility by 9
A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 9 if the sum of its digits is divisible by 9. Every number divisible by 9 is also divisible by 3, but not vice versa.
Divisibility by 24
Multiple of 4 that is also a multiple of 6
Multiple of 24
A number divisible by both 4 and 6 is not necessarily divisible by 24 (e.g., 12 is divisible by 4 and 6 but not by 24). A number divisible by 24 must be divisible by both 3 and 8 (since 3 and 8 are coprime and their product is 24).
Exam Focus
Marks pattern: 2-3 MCQs (1 mark each) on divisibility rules and parity, 1 short answer (2-3 marks) on sum of consecutive numbers or digital root, 1 long answer (4-5 marks) on cryptarithm or proving a divisibility property using algebra.
Most asked: Divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, 11 · Finding remainders using digit sum (divisibility by 3 and 9) · Properties of even and odd numbers (parity) · Sum of consecutive numbers · Digital root and its relation to divisibility by 9
Don't skip: Cryptarithm puzzles (5 marks) — requires logical reasoning and systematic trial.; Proving that the sum of two multiples of a number is also a multiple (using algebra) — often appears as a 3-4 mark question.; Finding numbers with given remainders (e.g., 'Find a number that leaves remainder 2 when divided by 3 and remainder 3 when divided by 4') — 4 marks.
One-Line Revision (read aloud before the exam)
  1. All odd numbers can be written as the sum of two consecutive numbers.
  2. The sum of any four consecutive numbers with + and - signs in all eight ways always gives an even number.
  3. An even number can be written as 2k, and an odd number as 2k + 1, where k is any integer.
  4. A number is divisible by 2 if its units digit is 0, 2, 4, 6, or 8.
  5. A number is divisible by 3 if the sum of its digits is divisible by 3.
  6. A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
  7. A number is divisible by 5 if its units digit is 0 or 5.
  8. A number is divisible by 6 if it is divisible by both 2 and 3.
  9. A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
  10. A number is divisible by 9 if the sum of its digits is divisible by 9.
  11. A number is divisible by 10 if its units digit is 0.
  12. A number is divisible by 11 if the difference between the sum of digits in odd places and even places is 0 or a multiple of 11.
  13. The digital root of a number is the single digit obtained by repeatedly adding its digits.
  14. The digital root of a multiple of 9 is always 9.
  15. If a divides M and a divides N, then a divides M + N and M - N.
  16. If a number is divisible by k, it is divisible by all factors of k, but not necessarily by multiples of k.
  17. To check divisibility by 24, check divisibility by 3 and 8 (since they are coprime).
  18. In a cryptarithm, each letter stands for a unique digit, and the first digit of a number cannot be 0.
  19. Numbers that leave remainder r when divided by d can be written as d × k + r.
  20. The sum of two even numbers is always even, but it is a multiple of 4 only under specific conditions.

MCQs — 10 Questions with Answers

Q1. When 4 consecutive numbers are chosen and combined with '+' and '–' signs in any way, what can we always say about the resulting expressions?

  • A. All expressions will always have the same parity (all even or all odd) ✓
  • B. All expressions will always be positive numbers
  • C. All expressions will always be divisible by 4
  • D. All expressions will always equal zero

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?

  • A. All even numbers are multiples of 4
  • B. Even numbers can be classified into two types based on remainders when divided by 4: those with remainder 0 and those with remainder 2 ✓
  • C. Even numbers always leave a remainder of 1 when divided by 2
  • D. Even numbers cannot be written as the sum of two consecutive numbers

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?

  • A. odd + odd = even
  • B. even + even = even
  • C. odd + even = even ✓
  • D. odd – odd = even

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?

  • A. The expression is always odd because 4m is even and 2n is even
  • B. The expression is always even because it can be written as 2(2m + n), showing 2 is a factor ✓
  • C. The expression is always divisible by 8
  • D. The expression is sometimes even and sometimes odd depending on the values of m and n

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?

  • A. It changes by 2b, which is always even ✓
  • B. It changes by b, which could be even or odd
  • C. It changes by –2b, which is always even
  • D. It changes by 4b, which depends on the value of b

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?

  • A. 672 – 348 (even – even)
  • B. 43 + 37 (odd + odd) ✓
  • C. 119 × 303 (odd × odd)
  • D. 809 + 214 (odd + 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?

  • A. 2u – 4v for any integers u and v
  • B. 4k × 3j for any integers k and j
  • C. x² + 2 for any integer x
  • D. 6m – 3n for any integers m and n ✓

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?

  • A. The sum will always be a multiple of 4
  • B. The sum will always leave a remainder of 2 when divided by 4
  • C. The sum will always leave a remainder of 2 when divided by 4 (similar to adding an even and an odd number to get an odd result) ✓
  • D. The sum will sometimes be a multiple of 4 and sometimes not

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?

  • A. When x is any positive integer
  • B. When x is an even integer
  • C. When x is an odd integer ✓
  • D. When x is divisible by 4

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?

  • A. Yes, because odd = even + odd, and any two consecutive numbers are one even and one odd ✓
  • B. No, because some odd numbers cannot be expressed this way
  • C. Yes, but only for odd numbers greater than 3
  • D. No, because the sum of two consecutive numbers is always even

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).

Flashcards

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.

Important Board Questions

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.

Assertion-Reason Questions for Number Play

Q1. Assertion (A):

The sum of any even number and any odd number is always an odd number.

Reason (R):

Even numbers have 2 as a factor while odd numbers do not, and when these are added, the result cannot have 2 as a factor.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains WHY the sum of an even and odd number is odd. The lack of a factor of 2 in the odd number prevents the sum from being even.

Q2. Assertion (A):

The expression 4m + 2n will always evaluate to an even number for any integer values of m and n.

Reason (R):

Both 4m and 2n are even numbers, and the sum of two even numbers is always even.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true (4m + 2n = 2(2m + n), so it has factor 2). The reason is also true (even + even = even). However, the reason is NOT the ONLY correct explanation. An equally valid reason is that 2 is a common factor of both terms, making the entire expression even. Both explanations work, but neither is THE unique correct explanation for this assertion.

Q3. Assertion (A):

The expression x² + 2 will always give an even number for any integer value of x.

Reason (R):

The number 2 is even, and adding it to any square will produce an even number.

Answer: D — A is false but R is true — The assertion is FALSE because if x is odd, then x² is odd, and odd + even = odd (e.g., 3² + 2 = 11). The reason is also false because it ignores the parity of x². The assertion fails even though the reason's logic seems plausible to a student who hasn't fully grasped that x² can be odd.

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Case-Based Questions (CBQ) for Number Play

Case Study:

A construction company is organizing wooden planks for a project. The manager assigns planks to three consecutive work stations: Station A gets 12 planks, Station B gets 13 planks, and Station C gets 14 planks. To test worker efficiency, the manager asks workers to calculate expressions using these three numbers with '+' and '–' signs. For example: 12 + 13 + 14 = 39, and 12 + 13 – 14 = 11. The manager notices that some workers correctly predict whether these expressions give even or odd results without calculating.

Q1. What is the parity of the expression 12 + 13 + 14?

  • A. Even
  • B. Odd ✓
  • C. Cannot be determined
  • D. Both even and odd

Answer: B

Q2. If the manager asks for 12 – 13 + 14, what parity will the result have?

  • A. Even
  • B. Odd ✓
  • C. Neither even nor odd
  • D. Depends on the order

Answer: B

Q3. Why can workers predict the parity of such expressions involving three consecutive numbers?

  • A. Because the parity depends only on the parities of the individual numbers and operations, not their actual values ✓
  • B. Because consecutive numbers always add to multiples of 3
  • C. Because three consecutive numbers always contain two even and one odd number
  • D. Because the expressions are always odd

Answer: A

Q4. Which expression will give an EVEN result?

  • A. 12 + 13 + 14
  • B. 12 – 13 + 14
  • C. 12 + 13 – 14
  • D. 12 – 13 – 14 ✓

Answer: D

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