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Question 4 - Figure it out - Page 132, 133, 134 - Chapter 5 Class 8 - Number Play (Ganita Prakash)
Last updated at January 8, 2026 by Teachoo
Figure it out - Page 132, 133, 134
Class 8 (Ganita Prakash - 1, 2 & Old NCERT)
Chapter 5 Class 8 - Number Play (Ganita Prakash)
- Sum of Consecutive Numbers
- Parity of Arithmetic & Algebraic Expressions
- Pairs to Make Fours
- Factors and Multiples
- Always, Sometimes, or Never
- What Remains?
- Figure it out - Page 122, 123
- Divisibility Rules - 2, 5, 10 and 3
- Shortcut for Divisibility by 9
- Figure it out - Page 126
- Shortcut for Divisibility by 11
- Divisibility Shortcuts for Other Numbers
- Digital Roots
- Figure it out - Page 131
- Digits in Disguise
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Figure it out - Page 132, 133, 134
Transcript
Question 4 (i) Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”. (i) Examine if her conjecture is true for any multiple of 9.A number is divisible by 9 if the sum of its digits is divisible by 9. If we reverse the digits, the sum remains the same So, the conjecture is true for any multiple of 9 Example: 18 and 81 are both divisible by 9 Question 4 (ii) Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”. (ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?A number is divisible by 9 if the sum of its digits is divisible by 9. If we move the digits in any other way, the sum will remain the same. And, if the sum remains the same, the number formed is still a multiple of 9. So, the answer is Yes, any shuffling is possible Example: 189 and 918 are both divisible by 9
