Questions

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Let the required number be 'X'

Condition 1 :  

$$\begin{gathered} XX = 2X \\ => 2X2 = X \\ => X = 12 \end{gathered}$$

Condition 2 :

$$\begin{gathered} X × X = X2 \\ => X = 12 \end{gathered}$$

Therefore, 12 is the number which becomes double when it is divided by itself and becomes half when it is multiplied by itself.

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All the above will stay where they are when they goes off. Basically anything that can swithch on or off will come under this.

Let the two digits be x & y

From given data in the question, we get

x + y = 3(x - y)

x + y = 3x - 3y

=> 2x = 4y

x/y = 2/1

For having the 2:1, numbers satisfying are

12, 21, 24, 42, 36, 63, 48, 84

From this 36 is eleminated because it is a perfect square and no others are prime or perfect squares.

Hence, there exists 7 such 2-digit numbers => 12, 21, 24, 42, 63, 48, 84.

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