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1. What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder ?

A. 312

B. 242

C. 1562

D. 1586

Solution:

\begin{aligned} LCM of (3, 5, 6, 8, 10, 12) \\ = 3 \times 5 \times 2 \times 4 \\ = 120 \\ Required number is \\ \frac{120 K + 2}{22} = \frac{10 K + 2}{22} \\ at K =  2, \frac{10 K + 2}{22} \\ \Rightarrow Remainder  =  0 \\ The given condition satisfied \\ = 120 K + 2 \\ = 240 + 2 \\ = 242 \end{aligned}

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