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1. The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case , but when divided by 7 leaves no remainder , is = ?

A. 189

B. 182

C. 175

D. 91

Solution:

LCM of (5, 10, 12, 15) = 5 × 2 × 6 = 60
smallest number divided by (5, 10, 12, 15)
leaves remainder 2 and when divided by 7 leaves no remainder is

\begin{aligned} = \frac{60 K + 2}{7} = \frac{4 K + 2}{7} \\ At K  =  3, \frac{4 K + 2}{7} \\ \Rightarrow remainder  =  0 \\ ∴ number  =  60K + 2 \\ =  60 \times 3 + 2 \\ = 182 \end{aligned}

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