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 and gives it back; Amitabh again doubles the received number. The first person whose result exceeds 1000 loses. Let $x$ be the smallest initial number that results in a win for Amitabh. The sum of the digits of $x$ is:

• A. 3
• B. 5
• C. 7
• D. 9
• E. None of these

Hint:

When a game alternates affine operations (like $2y$ and $2y+50$), identify the ``winning window'' for the losing move, then back-solve via the inverse transformation to find minimal feasible starts.

Answer 1

The Correct Option is C

Step 1: Model one round.
If Amitabh receives y, he outputs 2y. Then Sashi outputs 2y+50.
Amitabh wins iff Sashi is the first to exceed 1000.

Step 2: Characterize the “losing for Sashi” window.
On Sashi’s turn we need
2y ≤ 1000 and 2y+50 > 1000 ⇒ y ∈ (475, 500].
Call the value before Sashi’s losing move y_n.

Step 3: Work backwards to find the smallest initial x.
The value passed back to Amitabh each time satisfies
y_k+1 = 2y_k + 50 ⟺ y_k = (y_k+1 − 50)/2.
To minimize the initial x = y₁, take the smallest y_n ∈ {476,…,500} that allows repeated back-steps to stay integer and positive.

Try y_n = 478:
y_n−1 = (478 − 50)/2 = 214,
y_n−2 = (214 − 50)/2 = 82,
y_n−3 = (82 − 50)/2 = 16.

Thus an initial x = 16 produces the sequence
16 → 32 → 82 → 164 → 214 → 428 → 478 → 956 → 1006,
so Sashi exceeds 1000 first and loses.
Checking smaller options from y_n ∈ [476, 500] shows no back-stepped integer gives x < 16 and x ≥ 1. Hence the smallest winning start is

x = 16 ⇒ sum of digits = 1 + 6 = 7.

Final Answer: 7