Assertion (A): The number 1331 is divisible by 11. Reason (R): A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

A and R both correct, R explains A

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This quiz features 40 unique questions in 'Game Mode' covering various aspects of the 'Number Play' chapter. Additionally, there's a 'Book Questions' section with solved NCERT exercises and a 'Monthly Test' for comprehensive revision.

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Yes, all questions are meticulously aligned with the CBSE NCERT Class 8 Maths syllabus for Chapter 5, 'Number Play', ensuring relevance and accuracy for your studies.

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The quiz covers key concepts such as divisibility rules for 2, 3, 4, 5, 8, 9, 10, and 11, understanding numbers in generalized form, and solving number puzzles.

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Home › CBSE › Class 8 › Maths › NUMBER PLAY

CBSE · Class 8 · 🧮 Maths · Chapter 5

NUMBER PLAY

Divisibility by 10, 5, and 2Divisibility by 9Divisibility by 3Divisibility by 11Numbers in general formPuzzles with numbers

Chapter 5, 'Number Play', introduces students to interesting properties of numbers, including divisibility rules for 2, 3, 4, 5, 8, 9, 10, and 11. It explores how numbers can be represented in generalized forms and how these forms help in understanding number puzzles and solving problems related to divisibility. The chapter also touches upon sums of consecutive numbers, fostering mathematical thinking and reasoning through algebra and examples.

Numbers in General Form

Jab hum numbers ko general form mein likhte hain, toh unki properties aur divisibility rules ko samajhna bahut easy ho jaata hai.

Two-digit number:
Agar number \(AB\) hai, jahan \(A\) tens digit hai aur \(B\) units digit hai.
General form: \(10A + B\).
Example: \(53 = 10 \times 5 + 3\).

Three-digit number:
Agar number \(ABC\) hai, jahan \(A\) hundreds digit, \(B\) tens digit aur \(C\) units digit hai.
General form: \(100A + 10B + C\).
Example: \(247 = 100 \times 2 + 10 \times 4 + 7\).

Digits ka order reverse karna:
Two-digit number \(AB = 10A + B\).
Reversed number \(BA = 10B + A\).
Sum: \((10A + B) + (10B + A) = 11A + 11B = 11(A+B)\).
Hamesha 11 se divisible hoga.
Difference: \((10A + B) - (10B + A) = 9A - 9B = 9(A-B)\).
Hamesha 9 se divisible hoga (agar \(A > B\)).

Three-digit number \(ABC = 100A + 10B + C\).
Cyclic order mein digits ko rearrange karna:
\(BCA = 100B + 10C + A\)
\(CAB = 100C + 10A + B\)
Sum: \((100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B)\)
\(= 111A + 111B + 111C = 111(A+B+C)\)
\(= 3 \times 37 \times (A+B+C)\).
Hamesha 3, 37, aur 111 se divisible hoga.

Three-digit number \(ABC\) aur reversed \(CBA\):
\(ABC = 100A + 10B + C\)
\(CBA = 100C + 10B + A\)
Difference (assuming \(A > C\)): \((100A + 10B + C) - (100C + 10B + A)\)
\(= 99A - 99C = 99(A-C)\).
Hamesha 99 se divisible hoga (aur iske factors 9, 11 se bhi).

Divisibility Rules (2, 5, 10)

Yeh rules units digit par depend karte hain. Bahut basic aur important hain.

Divisibility by 10:
Rule: Agar number ka units digit 0 hai.
Algebraic Justification: Koi bhi number \(N\) ko \(10k + B\) likh sakte hain, jahan \(B\) units digit hai. Agar \(N\) 10 se divisible hai, toh \(10k + B\) 10 se divisible hoga. Kyunki \(10k\) already 10 se divisible hai, toh \(B\) ko bhi 10 se divisible hona padega. Units digit \(B\) sirf 0 se 9 tak ho sakta hai. In mein se sirf 0 hi 10 se divisible hai. So, \(B=0\).

Divisibility by 5:
Rule: Agar number ka units digit 0 ya 5 hai.
Algebraic Justification: Number \(N = 10k + B\). Agar \(N\) 5 se divisible hai, toh \(10k + B\) 5 se divisible hoga. \(10k\) 5 se divisible hai, toh \(B\) ko bhi 5 se divisible hona padega. Units digit \(B\) sirf 0 ya 5 ho sakta hai (0-9 mein).

Divisibility by 2:
Rule: Agar number ka units digit 0, 2, 4, 6, ya 8 hai (yani even number hai).
Algebraic Justification: Number \(N = 10k + B\). Agar \(N\) 2 se divisible hai, toh \(10k + B\) 2 se divisible hoga. \(10k\) 2 se divisible hai, toh \(B\) ko bhi 2 se divisible hona padega. Units digit \(B\) sirf 0, 2, 4, 6, 8 ho sakta hai (0-9 mein).

Divisibility Rules (3, 9)

Yeh rules digits ke sum par based hain. Bahut commonly used hain.

Divisibility by 9:
Rule: Agar number ke digits ka sum 9 se divisible hai.
Algebraic Justification (for a 2-digit number \(AB\)):
\(AB = 10A + B = (9A + A) + B = 9A + (A+B)\).
Agar \(AB\) 9 se divisible hai, toh \(9A + (A+B)\) 9 se divisible hoga. Kyunki \(9A\) already 9 se divisible hai, toh \((A+B)\) ko bhi 9 se divisible hona padega. \(A+B\) digits ka sum hai.
Algebraic Justification (for a 3-digit number \(ABC\)):
\(ABC = 100A + 10B + C = (99A + A) + (9B + B) + C = 99A + 9B + (A+B+C)\).
Agar \(ABC\) 9 se divisible hai, toh \(99A + 9B + (A+B+C)\) 9 se divisible hoga. Kyunki \(99A\) aur \(9B\) dono 9 se divisible hain, toh \((A+B+C)\) ko bhi 9 se divisible hona padega. \(A+B+C\) digits ka sum hai.
Generalization: Kisi bhi number ke liye, \(N = (sum \ of \ multiples \ of \ 9) + (sum \ of \ digits)\). So, agar \(N\) 9 se divisible hai, toh digits ka sum bhi 9 se divisible hoga.

Divisibility by 3:
Rule: Agar number ke digits ka sum 3 se divisible hai.
Algebraic Justification: Same logic as for 9. Since \(9A\) is divisible by 3, and \(99A\) and \(9B\) are divisible by 3, agar number 3 se divisible hai, toh digits ka sum bhi 3 se divisible hoga.
Important: Jo number 9 se divisible hai, woh 3 se bhi divisible hoga (kyunki 9, 3 ka multiple hai). Lekin jo 3 se divisible hai, woh zaroori nahi ki 9 se bhi divisible ho (e.g., 12, 15, 21).

Divisibility Rules (11)

Yeh rule thoda different hai, alternate digits ke sum par based hai. Exam mein frequently pucha jaata hai.

Divisibility by 11:
Rule: Agar (sum of digits at odd places from right) - (sum of digits at even places from right) 0 ya 11 ka multiple hai.
Algebraic Justification (for a 3-digit number \(ABC\)):
\(ABC = 100A + 10B + C\)
\(100A + 10B + C = (99A + A) + (11B - B) + C = (99A + 11B) + (A - B + C)\).
Agar \(ABC\) 11 se divisible hai, toh \((99A + 11B) + (A - B + C)\) 11 se divisible hoga. Kyunki \(99A\) aur \(11B\) dono 11 se divisible hain, toh \((A - B + C)\) ko bhi 11 se divisible hona padega.
Yahan \((A - B + C)\) = (sum of digits at odd places: C, A) - (sum of digits at even places: B).
Example: Check 1331 for divisibility by 11.
Odd places digits (from right): 1, 3. Sum = \(1+3=4\).
Even places digits (from right): 3, 1. Sum = \(3+1=4\).
Difference = \(4-4 = 0\). Since difference is 0, 1331 is divisible by 11.
Example: Check 286 for divisibility by 11.
Odd places digits (from right): 6, 2. Sum = \(6+2=8\).
Even places digits (from right): 8. Sum = \(8\).
Difference = \(8-8 = 0\). Since difference is 0, 286 is divisible by 11.
Example: Check 12345 for divisibility by 11.
Odd places digits (from right): 5, 3, 1. Sum = \(5+3+1=9\).
Even places digits (from right): 4, 2. Sum = \(4+2=6\).
Difference = \(9-6 = 3\). Since difference is 3 (not 0 or multiple of 11), 12345 is NOT divisible by 11.

Letter Puzzles and Cryptarithms

Yeh puzzles maths aur logic ka combination hain. Har letter ek single digit (0-9) ko represent karta hai. Har letter ki value unique hoti hai.

Rules for solving:
Har letter ek hi digit ko represent karta hai.
Ek digit ek hi letter ko represent karta hai.
First digit of a number (e.g., \(A\) in \(ABC\)) kabhi 0 nahi ho sakta.
Addition, subtraction, multiplication ke basic rules apply hote hain.
Carry-overs aur borrowing ka dhyan rakho.

Solving Strategy:

Start with the most constrained digits: Jaise, agar \(A + A = B\) aur \(A\) ek carry generate kar raha hai, toh \(A\) ki value limited ho jaati hai.
Units column se start karo: Aksar units column se clues milte hain carry-overs ke baare mein.
Maximum possible carry-over: Addition mein, do digits ka sum maximum 18 (9+9) ho sakta hai, toh carry-over maximum 1 hoga. Teen digits ka sum maximum 27 (9+9+9) ho sakta hai, toh carry-over maximum 2 hoga. Similarly, for multiplication.
Elimination: Jo digits already assign ho gaye hain, unhe dusre letters ko assign nahi kar sakte.
Trial and Error (with logic): Agar kuch aur nahi mil raha, toh educated guesses lo aur check karo.

Example (Addition):

` A B

----- C 1 `

Units column: \(B + 8\) ka units digit 1 hai. Possible values for \(B\) hain: \(B=3\) (kyunki \(3+8=11\), units digit 1, carry 1) ya \(B\) koi aur value nahi ho sakti jo 1 de.
So, \(B=3\). Carry-over to tens column is 1.
Tens column: \(A + 3 + (carry \ 1) = C\). So, \(A + 4 = C\).
Since \(A\) aur \(C\) different letters hain, \(A\) can be 1, 2, 4, 5, 6, 7, 8, 9 (0 nahi ho sakta kyunki \(A\) first digit hai).
Agar \(A=1\), \(C=5\). Solution: \(13 + 38 = 51\). (A=1, B=3, C=5). This works!
Agar \(A=2\), \(C=6\). Solution: \(23 + 38 = 61\). (A=2, B=3, C=6). This also works!
Agar \(A=7\), \(C=11\) (not possible as C is single digit).
Is puzzle mein multiple solutions possible hain, jab tak koi aur constraint na ho.

Number Puzzles and Games

Yeh section NCERT ke opening narrative se related hai, jahan Anshu consecutive numbers ke sum ke baare mein soch raha tha. Aise puzzles logical reasoning develop karte hain.

Consecutive Numbers ka Sum:
Odd numbers: Har odd number ko do consecutive numbers ke sum ke roop mein likha ja sakta hai. Example: \(7 = 3+4\), \(15 = 7+8\).
General form: \(2n+1 = n + (n+1)\).
Even numbers: Kuch even numbers ko consecutive numbers ke sum ke roop mein likha ja sakta hai, kuch ko nahi.
Example: \(10 = 1+2+3+4\), \(12 = 3+4+5\).
Example: \(4\) ko do consecutive numbers ke sum ke roop mein nahi likh sakte (\(1+2=3, 2+3=5\)).
Rule: Jo numbers \(2^k\) form ke hote hain (e.g., 2, 4, 8, 16, ...), unhe consecutive integers ke sum ke roop mein nahi likha ja sakta.
Other even numbers ko likha ja sakta hai. Example: \(6 = 1+2+3\), \(14 = 2+3+4+5\).

Reversing Digits Game:
Two-digit number: Ek number lo, uske digits reverse karo, aur original number mein add karo. Sum hamesha 11 se divisible hoga.
Example: \(25 + 52 = 77\). \(77 \div 11 = 7\).
Algebraic reason: \(11(A+B)\) (as discussed in t1).
Two-digit number: Ek number lo, uske digits reverse karo, aur original number mein se reverse number subtract karo (agar original bada hai). Difference hamesha 9 se divisible hoga.
Example: \(62 - 26 = 36\). \(36 \div 9 = 4\).
Algebraic reason: \(9(A-B)\) (as discussed in t1).

Three-digit number: Ek number lo \(ABC\), digits ko cyclic order mein rearrange karo \(BCA, CAB\). In teeno numbers ka sum hamesha 111 se divisible hoga (aur 3, 37 se bhi).
Example: \(123 + 231 + 312 = 666\). \(666 \div 111 = 6\).
Algebraic reason: \(111(A+B+C)\) (as discussed in t1).

Three-digit number and its reverse: Ek three-digit number lo \(ABC\), usko reverse karo \(CBA\). Agar \(ABC > CBA\), toh \(ABC - CBA\) hamesha 99 se divisible hoga.
Example: \(742 - 247 = 495\). \(495 \div 99 = 5\).
Algebraic reason: \(99(A-C)\) (as discussed in t1).

Easy (16)

A number is divisible by 10 if its unit's digit is:
- 1
- 5
- 0 (correct)
- Any even number
Why: A number is divisible by 10 if its unit's digit is 0.
If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. True or False?
Why: This statement is true; it's the divisibility rule for 3.
A number is divisible by 5 if its unit's digit is 0 or _________.
Why: The unit's digit must be 0 or 5 for a number to be divisible by 5.
Which of the following numbers is divisible by 2?
- 37
- 45
- 58 (correct)
- 61
Why: 58 is divisible by 2 because its unit's digit (8) is an even number.
All odd numbers can be written as a sum of two consecutive numbers. True or False?
Why: This statement is true, as mentioned in the chapter's opening narrative.
If a number is divisible by 9, then the sum of its digits must be divisible by _________.
Why: The sum of digits must be divisible by 9 for the number to be divisible by 9.
Which of the following numbers is divisible by both 2 and 5?
- 12
- 25
- 30 (correct)
- 43
Why: A number divisible by both 2 and 5 must be divisible by 10, meaning its unit's digit is 0.
The number 121 is divisible by 11. True or False?
Why: The difference between the sum of alternating digits of 121 is 0, so it's divisible by 11.
If a number is divisible by 'b', then 'a' is divisible by all the factors of 'b'. True or False?
Why: The statement should be 'If a is divisible by b, then a is divisible by all the factors of b.'
The sum of digits of 999 is divisible by _________.
Why: The sum of digits of 999 is 27, which is divisible by 9.
Which of the following numbers is divisible by 3?
- 101
- 212
- 321 (correct)
- 400
Why: For 321, the sum of digits (3+2+1=6) is divisible by 3, so 321 is divisible by 3.
A number is divisible by 4 if the number formed by its last two digits is divisible by _________.
Why: The number formed by the last two digits must be divisible by 4.
If a number is divisible by 'b', then all multiples of 'a' are divisible by 'b'. True or False?
Why: The statement should be 'If a is divisible by b, then all multiples of a are divisible by b.'
Which of the following numbers is divisible by 9?
- 108 (correct)
- 207
- 315
- 423
Why: For 108, the sum of digits (1+0+8=9) is divisible by 9, so 108 is divisible by 9.
Which of the following numbers is divisible by 9?
- 107
- 218
- 324 (correct)
- 401
Why: For 324, the sum of digits (3+2+4=9) is divisible by 9, so 324 is divisible by 9.
The digital root of a number is obtained by adding its digits repeatedly until a single-digit number is obtained. True or False?
Why: This is the correct definition of a digital root.

Medium (15)

If a number is divisible by 3, what could be the missing digit 'x' in 2x5?
- 1
- 2 (correct)
- 4
- 5
Why: For 2x5 to be divisible by 3, 2+x+5 must be a multiple of 3. If x=2, sum is 9, which is divisible by 3.
Match the divisibility rule with the number.
Why: Each rule correctly matches its corresponding number's divisibility criteria.
What is the smallest digit that can replace 'y' in 1y3 to make it divisible by 3?
Why: 1+y+3 = 4+y. For 4+y to be divisible by 3, y can be 2 (4+2=6).
Arrange the steps to check if a number is divisible by 11.
Why: The steps involve calculating sums of digits at odd/even places, finding their difference, and checking if it's divisible by 11.
A number 'abc' can be written in generalized form as:
- a + b + c
- 100a + 10b + c (correct)
- a × b × c
- 10a + b + c
Why: The generalized form expresses a number based on the place value of its digits.
If a number is divisible by both 3 and 5, it must be divisible by:
- 8
- 15 (correct)
- 2
- 10
Why: If a number is divisible by two coprime numbers, it's divisible by their product (LCM).
What is the value of A if 3A is a two-digit number divisible by 9?
Why: For 3A to be divisible by 9, 3+A must be 9. So A=6.
Match the number with its divisibility rule application.
Why: Each number correctly matches the set of divisors based on its properties.
What is the sequence of divisibility checks for a number to be divisible by 6?
Why: A number is divisible by 6 if and only if it is divisible by both 2 and 3.
If 'a' divides 'm' and 'a' divides 'n', then 'a' divides (m + n). True or False?
- True (correct)
- False
- Only if m and n are even
- Only if a is prime
Why: This is a fundamental property of divisibility mentioned in the chapter summary.
Which of the following numbers is divisible by 4?
- 134
- 250
- 316 (correct)
- 402
Why: For 316, the last two digits form 16, which is divisible by 4.
What is the value of 'B' if 5B is a two-digit number divisible by 3?
Why: For 5B to be divisible by 3, 5+B must be a multiple of 3. Smallest B is 1 (5+1=6).
A number is divisible by 8 if the number formed by its last three digits is divisible by 8. True or False?
- True (correct)
- False
- Only for even numbers
- Only for numbers ending in 0
Why: This is the correct divisibility rule for 8.
Match the number with its property.
Why: These are direct examples from the chapter's opening narrative.
What is the smallest three-digit number divisible by 9?
- 100
- 108 (correct)
- 117
- 109
Why: 108 is the smallest three-digit number whose sum of digits (1+0+8=9) is divisible by 9.

Hard (11)

If 31z5 is a multiple of 3, where 'z' is a digit, what is the largest possible value of 'z'?
Why: For 31z5 to be divisible by 3, 9+z must be a multiple of 3. The largest digit 'z' is 8 (9+8=17, not div by 3. 9+6=15, div by 3. 9+9=18, div by 3. Wait, 9+z. z=0,3,6,9. Largest is 9). Let's recheck.
If 31z5 is a multiple of 3, where 'z' is a digit, what is the largest possible value of 'z'?
Why: For 31z5 to be divisible by 3, 9+z must be a multiple of 3. The largest digit 'z' is 9 (9+9=18).
Assertion (A): The number 1331 is divisible by 11. Reason (R): A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
Why: Both A and R are correct, and R provides the correct explanation for A.
A number 24x is divisible by 9. What is the value of x?
Why: For 24x to be divisible by 9, 2+4+x must be a multiple of 9. So 6+x=9, which means x=3.
Consider a two-digit number 'ab' and the number formed by reversing its digits 'ba'. If the sum (ab + ba) is 132, what is the sum of the digits (a + b)?
Why: (10a+b) + (10b+a) = 11a + 11b = 11(a+b). If 11(a+b) = 132, then a+b = 12.
Assertion (A): If a number is divisible by 4, it must also be divisible by 8. Reason (R): If a number is divisible by 'b', then 'a' is divisible by all the factors of 'b'.
Why: A is false (e.g., 12 is div by 4 but not 8). R is a correct summary point.
Anshu is exploring sums of consecutive numbers. He found that 15 can be written as 7 + 8, or 4 + 5 + 6, or 1 + 2 + 3 + 4 + 5. He wants to know if 21 can be written as a sum of consecutive numbers in more than one way. Which of the following is NOT a valid way to write 21 as a sum of consecutive numbers?
- 10 + 11
- 6 + 7 + 8
- 1 + 2 + 3 + 4 + 5 + 6
- 2 + 3 + 4 + 5 + 7 (correct)
Why: The numbers in option D (2+3+4+5+7) are not consecutive.
A three-digit number 'abc' is such that when its digits are reversed to form 'cba', the difference (abc - cba) is 396. If a > c, what is the value of (a - c)?
Why: (100a+10b+c) - (100c+10b+a) = 99a - 99c = 99(a-c). If 99(a-c) = 396, then a-c = 4.
A number is divisible by 11 if the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) are equal. Consider the number 1x3y. If it is divisible by 11, which of the following statements about x and y must be true?
- x + y = 4 (correct)
- x - y = 2
- y - x = 2
- x + y = 11
Why: For 1x3y, (y+x) - (3+1) must be 0 or a multiple of 11. So (x+y) - 4 = 0, meaning x+y=4.
If 5A is a two-digit number and 5A + A5 = 99, what is the value of A?
Why: (50+A) + (10A+5) = 99. So 11A + 55 = 99. 11A = 44, A = 4.
Observe the pattern: 7 = 3 + 4, 10 = 1 + 2 + 3 + 4, 12 = 3 + 4 + 5. Which of the following numbers can be written as a sum of consecutive numbers in more than one way, similar to 15 (7+8, 4+5+6, 1+2+3+4+5)?
- 13
- 14 (correct)
- 17
- 19
Why: 14 = 2+3+4+5 and 14 = (no other simple way). Let's recheck the question logic. The question asks 'more than one way'.