Question

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:

Accepted Answer

C. Step 1: A gives $2x$, S makes $2x+50$. Step 2: A makes $4x+100$, S makes $4x+150$. Step 3: A makes $8x+300$, S makes $8x+350$. Step 4: A makes $16x+700$, S makes $16x+750$. Step 5: A makes $32x+1500$, S makes $32x+1550$. For Amitabh to win, Step 4 must end with Sashi exceeding 1000, while Amitabh at Step 4 is $\le 1000$. $16x + 750 > 1000 \implies 16x > 250 \implies x \ge 16$. For $x = 16$, Amitabh's result at step 4 is $16(16) + 700 = 956 \le 1000$, and Sashi's is $956 + 50 = 1006 > 1000$ (Sashi loses, Amitabh wins). Smallest initial $x = 16$. Sum of digits $= 1 + 6 = 7$.

Number Game Between Amitabh and Sashi

Explanation

Practise this under exam conditions