CG Board Class 9 Mathematics Chapter 6 Solutions: Playing with Numbers (संख्याओं में भी है खेल) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 6 Solutions: Playing with Numbers (संख्याओं में भी है खेल) in Bilingual Format (English & Hindi Medium).

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Q1

If A 3 B 4 C 5 D…


📝 Question:

If A = 3, B = 4, C = 5, D = 0 and each digit has to be used only one time then using these digits

💡 Solution:

(i) What will be the largest number?*

Arranging the given digits in descending order gives 5, 4, 3, 0. Therefore, the largest number is 5430 (or CBAD).

  • *(ii) Which number is bigger in ABCD and ADBC?

ABCD = 3450 and ADBC = 3045. Comparing the two numbers, 3450 > 3045. Thus, ABCD is bigger.

  • *(iii) What is the smallest number? Is it a three digit or four digit number?

Arranging in ascending order gives 0, 3, 4, 5. So DABC = 0345 = 345. Since a number starting with 0 is read as 345, it is a three-digit number.

  • *(iv) What is the value of DBAC? How many digits will be there in this number?

Substituting D = 0, B = 4, A = 3, C = 5 gives DBAC = 0435 = 435. It contains 3 digits.

Question 2

If A 1 B 2 C 3 D…


📝 Question:

If A = 1, B = 2, C = 3, D = 4 then

💡 Solution:

(i) Find the value of AB CD.

  • *(ii) What will be the value of AB + CD?

Q3

AB BA 77 then what is the value…


📝 Question:

AB + BA = 77, then what is the value of A, B? *Step-by-Step Solution /

💡 Solution:

Expanded form: (10A + B) + (10B + A) = 77 -> 11(A + B) = 77 -> A + B = 7. Since A and B are single non-zero digits, the possible pairs for (A, B) are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).

Question 2

Similarly make some more questions which can have…


📝 Question:

Similarly make some more questions which can have more than one answer

💡 Solution:

  • English Example: AB + BA = 99. Here 11(A + B) = 99 -> A + B = 9. Possible digit pairs for (A, B) are (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1).
  • : AB + BA = 99I 11(A + B) = 99 -> A + B = 9I (A, B) (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1)

Q5

Find out the value of B if 1B…


📝 Question:

Find out the value of B if 1B B = 96. **Step-by-Step Solution /

💡 Solution:

(10 + B) B = 96 -> B2 + 10B - 96 = 0 -> (B + 16)(B - 6) = 0. Since B is a positive digit, B = 6. Check: 16 6 = 96.

Question 2

Find out the value of M and N…


📝 Question:

Find out the value of M and N in 73M / 8 = 9N. *Step-by-Step Solution /

💡 Solution:

73M = 8 * (90 + N) -> 730 + M = 720 + 8N -> 10 + M = 8N. For single-digit M and N (0 to 9), if N = 2, then 10 + M = 16 -> M = 6. Check: 736 / 8 = 92. Thus, M = 6, N = 2.

Q7

Select any 3 digits without zero and do…


📝 Question:

/ :** Select any 3 digits (without zero) and do Neha’s activity. Did you find sum of digits as Alok did? ?)

💡 Solution:

Let selected digits be 1, 4, 8.

  • All two-digit numbers formed: 14, 48, 81, 41, 84, 18.
  • Sum = 14 + 48 + 81 + 41 + 84 + 18 = 286.
  • Dividing sum by 22: 286 / 22 = 13.
  • Sum of selected digits = 1 + 4 + 8 = 13. Yes, the quotient equals the sum of digits!

Q8

Take any two digit number Interchange its digits…


📝 Question:

Take any two digit number. Interchange its digits, add both numbers. Why is their sum always divisible by 11?

💡 Solution:

Let two-digit number = 10a + b. Reversed number = 10b + a.

  • Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
  • Since 11 is a factor, the sum is always divisible by 11.

Question 2

Think of a three digit number Reverse its…


📝 Question:

Think of a three digit number. Reverse its digits and subtract smaller from bigger. Is it a multiple of 99? Why?

💡 Solution:

Let original number = 100a + 10b + c (where a > c). Reversed number = 100c + 10b + a.

  • Difference = (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c).
  • Since 99 is a factor, the difference is always a multiple of 99.

Q10

Are all numbers that are divisible by 5…


📝 Question:

/ :** Are all numbers that are divisible by 5, also divisible by 10? Explain why

💡 Solution:

No. A number divisible by 5 can end in either 0 or 5, but a number divisible by 10 must end only in 0. For example, 15, 25, 35 are divisible by 5, but not by 10.

Q11

Check the divisibility of 373 644 343 861…


📝 Question:

/ :** Check the divisibility of 373, 644, 343, 861 by 7 using the rule 7(14a + b) + (2a + 3b + c)

💡 Solution:

(i) 373 (a = 3, b = 7, c = 3):** 2(3) + 3(7) + 3 = 6 + 21 + 3 = 30. 30 is not divisible by 7 -> 373 is NOT divisible by 7.

  • (ii) 644 (a = 6, b = 4, c = 4): 2(6) + 3(4) + 4 = 12 + 12 + 4 = 28. 28 is divisible by 7 -> 644 IS divisible by 7.
  • (iii) 343 (a = 3, b = 4, c = 3): 2(3) + 3(4) + 3 = 6 + 12 + 3 = 21. 21 is divisible by 7 -> 343 IS divisible by 7.
  • (iv) 861 (a = 8, b = 6, c = 1): 2(8) + 3(6) + 1 = 16 + 18 + 1 = 35. 35 is divisible by 7 -> 861 IS divisible by 7.

Q12

Check the divisibility of number 19151 by 11


📝 Question:

/ :** Check the divisibility of number 19151 by 11

💡 Solution:

  • Digits at odd places (from right): 1, 1, 1 -> Sum = 1 + 1 + 1 = 3.
  • Digits at even places (from right): 5, 9 -> Sum = 5 + 9 = 14.
  • Difference = 14 - 3 = 11.
  • Since 11 is divisible by 11, 19151 is divisible by 11.

Q13

Are all numbers that are divisible by 3…


📝 Question:

Are all numbers that are divisible by 3, also divisible by 9? Why or why not?

💡 Solution:

No. For example, 6, 12, 15 are divisible by 3 but not by 9. Because 3 is a factor of 9, all multiples of 9 are divisible by 3, but not all multiples of 3 are multiples of 9.

  • : “I 6, 12, 15 3 9 3, 9 , 9 3 , 3 9

Question 2

Choose any number completely divisible by 6 Divide…


📝 Question:

Choose any number completely divisible by 6. Divide it by 2 and 3. What can you say about divisibility rule of 6?

💡 Solution:

Let number = 18. 18 / 2 = 9 and 18 / 3 = 6. It is divisible by both 2 and 3. Rule: A number is divisible by 6 if it is divisible by both 2 (even unit digit) and 3 (sum of digits divisible by 3).

  • : = 18I 18 / 2 = 9 18 / 3 = 6I 2 3 : 6 2 3

Question 3

Using digits 2 5 4 and 7 make…


📝 Question:

Using digits 2, 5, 4 and 7, make all numbers completely divisible by 15 without actual division

💡 Solution:

  • Divisibility by 15 requires divisibility by both 3 and 5.
  • Sum of digits = 2 + 5 + 4 + 7 = 18 (divisible by 3). So any 4-digit combination using all four digits is divisible by 3.
  • To be divisible by 5, the unit digit must be 5.
  • Arranging remaining digits {2, 4, 7} gives 6 possible numbers: 2475, 2745, 4275, 4725, 7245, 7425.
  • : 15 3 5 = 18 5 5 {2, 4, 7} 6 : 2475, 2745, 4275, 4725, 7245, 7425I

Question 4

Find the smallest number that will be divisible…


📝 Question:

Find the smallest number that will be divisible by 7 and 11 both

💡 Solution:

LCM(7, 11) = 7 * 11 = 77.

### Section 2: Exercise 6.1 Complete Step-by-Step Solutions / 6.1

Find out value of letters A, B, X, Y, Z, L, M, N as used in following questions

Question 1

i 1i B A 1 2 B 2…


📝 Question:

(i) / 1(i) ` B A + 1 2 B ——- 2 2 5 ` Step-by-Step Solution / : In units column: A + B = 5 or 15. From options and hundreds place, B + 1 + carry = 2 => B = 9. Substituting B = 9: A + 9 = 15 -> A = 6. Check: 96 + 129 = 225. Values: A = 6, B = 9. Question 1(ii) / 1(ii) ` 3 X Y + Y Y 2 ——- 1 0 1 8 ` Step-by-Step Solution / : Units place: Y + 2 = 8 -> Y = 6. Tens place: X + Y = X + 6 = 11 (with carry 1) -> X = 5. Hundreds place: 3 + Y + 1 = 3 + 6 + 1 = 10. Values: X = 5, Y = 6. Question 1(iii) / 1(iii) ` M N 6 —– M L N ` Step-by-Step Solution / : (10M + N) 6 = 100M + 10L + N. Units digit: 6N ends in N, which holds for N = 8 (6 8 = 48, carry 4). Tens/Hundreds place: 6M + 4 = 10M + L. If M = 1, then 6(1) + 4 = 10 -> M = 1, L = 0. Check: 18 6 = 108. Values: L = 0, M = 1, N = 8. Question 1(iv) / 1(iv) ` 1 Z Z —– 7 Z ` Step-by-Step Solution / : (10 + Z) Z = 70 + Z -> 10Z + Z2 = 70 + Z -> Z2 + 9Z - 70 = 0. Factoring gives (Z + 14)(Z - 5) = 0 -> Z = 5. Check: 15 5 = 75. Value: Z = 5. Question 1(v) / 1(v) ` X X + Y Y Y + 6 ——- 4 6 1 ` Step-by-Step Solution / : Equation: 11X + 111Y + 6 = 461 -> 11X + 111Y = 455. Testing digit Y = 4: 111 4 = 444. 11X = 455 - 444 = 11 -> X = 1. Check: 11 + 444 + 6 = 461. Values: X = 1, Y = 4. Question 1(vi) / 1(vi) ` 2 P Q + P Q 1 ——- Q 1 8 ` Step-by-Step Solution / : Units place: Q + 1 = 8 -> Q = 7. Tens place: P + Q = P + 7 = 11 (carry 1) -> P = 4. Hundreds place: 2 + P + 1 = 2 + 4 + 1 = 7 = Q. Values: P = 4, Q = 7. Question 1(vii) / 1(vii) ` M L 6 —– L L L ` Step-by-Step Solution / : (10M + L) 6 = 111L -> 60M + 6L = 111L -> 60M = 105L -> 4M = 7L. Since M and L are single non-zero digits, M = 7 and L = 4. Check: 74 6 = 444. Values: M = 7, L = 4. Section 3: Exercise 6.2 Complete Step-by-Step Solutions / 6.2 Question 1 / 1 Which of the following numbers are multiples of 5 and 10: 316, 9560, 205, 311, 800, 7936. Step-by-Step Solution / : Multiples of 5: Numbers ending in 0 or 5 are 9560, 205, 800. Multiples of 10: Numbers ending in 0 are 9560, 800. Question 2 / 2 If a number that is divisible by 3, has 8 in its unit place, then what can be possible digits in tens place? Step-by-Step Solution / : Let tens digit = x. Number = 10x + 8. Sum of digits = x + 8 must be a multiple of 3. For single-digit x (1 to 9): If x = 1: 1 + 8 = 9 (divisible by 3) If x = 4: 4 + 8 = 12 (divisible by 3) If x = 7: 7 + 8 = 15 (divisible by 3) Possible digits in tens place are 1, 4, 7. Question 3 / 3 If number 35P is a multiple of 5 then find the value of P? Step-by-Step Solution / : A number is a multiple of 5 if its units digit is 0 or 5. Therefore, P = 0 or 5. Question 4 / 4 If 6A3B is a number that is divisible by 9, find the value of A and B? Step-by-Step Solution / : Sum of digits = 6 + A + 3 + B = 9 + A + B. For divisibility by 9, (9 + A + B) must be divisible by 9, which implies A + B = 0 or A + B = 9. Valid digit pairs (A, B) include (7, 2), (3, 6), (2, 7), (6, 3), (0, 9), (9, 0), (4, 5), (5, 4), (1, 8), (8, 1), (0, 0). Question 5 / 5 Check the divisibility by 7 for following numbers: (i) 672 (ii) 905 (iii) 2205 (iv) 9751 Step-by-Step Solution /

💡 Solution:

(i) 672:** 67 - 2(2) = 67 - 4 = 63. Since 63 is divisible by 7 (63 / 7 = 9), 672 IS divisible by 7.

  • (iii) 2205: 220 - 2(5) = 220 - 10 = 210. Since 210 is divisible by 7 (210 / 7 = 30), 2205 IS divisible by 7.
  • (i) 913: Odd sum = 3 + 9 = 12, Even sum = 1. Difference = 12 - 1 = 11 -> Divisible by 11.
  • (iii) 3729: Odd sum = 9 + 7 = 16, Even sum = 2 + 3 = 5. Difference = 16 - 5 = 11 -> Divisible by 11.
  • (i) 169: 16 + 4(9) = 16 + 36 = 52. Since 52 is divisible by 13 (52 / 13 = 4), 169 IS divisible by 13.
  • (iii) 3146: 314 + 4(6) = 314 + 24 = 338. Step 2: 33 + 4(8) = 33 + 32 = 65. Since 65 is divisible by 13 (65 / 13 = 5), 3146 IS divisible by 13.

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