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Question 6 - Figure it out - Page 122, 123 - Chapter 5 Class 8 - Number Play (Ganita Prakash)
Last updated at January 7, 2026 by Teachoo
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?
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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
- Figure it out - Page 132, 133, 134
Transcript
Question 6 Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?A number which leaves remainder 2 when divided by 6 will be Number = Multiple of 6 + 2 = 6a + 2 Three such numbers would be 6a + 2 6b + 2 6c + 2 Now, adding these numbers Sum = (6a + 2) + (6b + 2) + (6c + 2) = (6a + 6b + 6c) + (2 + 2 + 2) = 6a + 6b + 6c + 6 = 6 (a + b + c + 1) Since the entire expression is multiplied by 6, the sum is always a multiple of 6. Thus, Tathagat’s claim is true.
