Question 57

Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each

time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let ‘x’ be the smallest initial number that results in a win for Amitabh. The sum of the digits of ‘x’ is:

Solution

Let the no. chosen by Amitabh be x

In the first step, the result got by Amitabh is 2x and the result got by Shashi is 2x+50.

In the second step, the result got by Amitabh is 4x+100 and the result got by Shashi is 4x+150.

In the third step, the result got by Amitabh is 8x+300 and the result got by Shashi is 8x+350.

In the fourth step, the result got by Amitabh is 16x+700 and the result got by Shashi is 16x+750.

In the fifth step, the result got by Amitabh is 32x+1500 and the result got by Shashi is 32x+1550.

In step 5, since both of their results are greater than 1000, so step 4 will be the last step.

Since Amitabh wins the game the result of Shashi must be greater than 1000.

Now, $$16x+750\ge\ 1000$$

or $$x\ge\ 16$$

Hence the minimum value of x should be 16 and the its sum of digits is 7.

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