Find the smallest number which when divided by $35, 56$ and $91$ leaves a remainder of $7$ in each case.
Step-by-Step Solution
Key Concept: Required number is $\text{LCM}(35, 56, 91) + 7$.
$35 = 5 \times 7$, $56 = 2^3 \times 7$, $91 = 7 \times 13$. [0.5 Mark]
$\text{LCM} = 2^3 \times 5 \times 7 \times 13 = 8 \times 5 \times 7 \times 13 = 3640$. [1.0 Mark]
Required number $= 3640 + 7 = 3647$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Prime factorisation: 0.5 Mark
LCM $= 3640$: 1.0 Mark
Adding remainder to get 3647: 0.5 Mark
Correct Answer: