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Question 2 - 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 2 โI take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8โ, claims Snehal. Examine his claim and justify your conclusion.Now, there are two numbers Number ๐(๐ต_๐ ) : Leaves remainder 8 when divided by 12 . Example: 20 (12 ร 1 + 8) Number ๐(๐ต_๐ ) : 4 short of a multiple of 12 . Example: 8 (12 ร 1 โ 4) Sum: 20+8=28. Check: Is 28 a multiple of 8 ? No Hence, we can say that Snehal's claim is False. The sum will be a multiple of 4 , but not necessarily a multiple of 8
