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Question 8 - Figure it out - Page 63, 64 - Chapter 3 Class 7 - Finding Common Ground (Ganita Prakash II)
Last updated at January 23, 2026 by Teachoo
Figure it out - Page 63, 64
Class 7 (Ganita Prakash 1, 2 & old NCERT)
Chapter 3 Class 7 - Finding Common Ground (Ganita Prakash II)
- The Greatest of All
- Prime Factorisation
- Finding HCF using Prime Factorisation
- Figure it out - Page 54
- Least, but not Last!
- Finding LCM using Prime Factorisation
- Patterns, Properties, and a Pretty Procedure!
- Figure it out - Page 59
- Miscellenaous Questions on HCF, LCM
- Efficient Procedures for HCF and LCM
- Property Involving both the HCF and the LCM
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Figure it out - Page 63, 64
Transcript
Question 8 Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.First, let’s find t smallest number that is divisible by 3, 4, 5 and 7 Now, Smallest number that is divisible by 3, 4, 5 and 7 = LCM of 3, 4, 5 & 7 Finding LCM using Shortcut Method There are no common factors between 3, 4 5, 7 ∴ LCM of 3, 4, 5, 7 = 3 × 4 × 5 × 7 = 12 × 35 = 420 Now, we need to find Smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11 Thus, the numbers would be multiples of 420 Multiples of 420 = 420, 840, 1260, 1680, 2100... Checking remaining of multiples when divided by 11 420÷11→ Remainder 2 840÷11→ Remainder 4 1260÷11→ Remainder 6 1680÷11→ Remainder 8 2100÷11→ Remainder 10 Thus, answer is 2100
