Question

In OperationSindoor, to destroy Pakistani Air Bases, India had three missile system options – SCALP (S), HAMMER (H) or BrahMos(B) (Excatly one system is to be used in entire operation)

Now at a time one Battery of missiles is fired (Each fired missile targets different Air Base) Now number of missiles per Battery in S,H and B is respectively 10,7 & 3.Batteries are fired till all Airbases are either destroyed or Number of undestroyed bases are less than number of Missiles fired in a battery.

It is known that number of undestroyed Pakistani Air Bases would have been 0,3 or 2 respectively according to which of S,H or B is used .Let minimum possible number of Pakistani Air bases is 'x' then sum of digits of x is :

Answer 1

8

Answer 2

Let $N$ be the initial number of Pakistani Air Bases. Let $n$ be the number of missiles per battery for a given missile system. The operation fires batteries of $n$ missiles until the number of undestroyed bases is less than $n$ or is 0. Let $U$ be the final number of undestroyed bases.

According to the rule, the operation stops when the number of undestroyed bases is less than $n$. Let $N_{rem}$ be the number of undestroyed bases at some point. If $N_{rem} \ge n$, a battery is fired, destroying $n$ bases (assuming each missile targets a different base). The new number of undestroyed bases is $N_{rem} - n$. This process continues until the number of undestroyed bases is less than $n$. Starting with $N$ bases, after firing $k$ batteries, the number of undestroyed bases is $N - kn$. The operation stops after $k$ batteries if $N - (k-1)n \ge n$ (so the $k$-th battery is fired) and $N - kn < n$ (so the operation stops after the $k$-th battery). Let $U = N - kn$ be the final number of undestroyed bases. The condition $N - kn < n$ is $U < n$. The condition $N - (k-1)n \ge n$ becomes $(N-kn) + n \ge n$, which is $U + n \ge n$, or $U \ge 0$. This is always true since $U$ is a number of bases.

So, the final number of undestroyed bases $U$ is the remainder when $N$ is divided by $n$, provided that the number of bases before the last step was at least $n$. Let $N = qn + r$, where $0 \le r < n$. If $q \ge 1$, then $N = (q-1)n + n + r$. After $q-1$ batteries, $n+r$ bases remain. Since $n+r \ge n$ (as $n \ge 3$ and $r \ge 0$), the $q$-th battery is fired. After the $q$-th battery, $r$ bases remain. Since $r < n$, the operation stops. So $U=r$. If $q = 0$, then $N = r$, and $0 \le r < n$. In this case, no batteries are fired as the initial number of bases is already less than $n$. So $U=N=r$. In both cases, $U$ is the remainder when $N$ is divided by $n$, i.e., $N \equiv U \pmod{n}$, and $0 \le U < n$.

We are given the values of $n$ and $U$ for the three missile systems:

1. SCALP (S): $n_S = 10$, $U_S = 0$. $N \equiv 0 \pmod{10}$. This means $N$ is a multiple of 10. $N = 10k_S$ for some integer $k_S \ge 0$. Since $U_S=0$, the operation stopped because all bases were destroyed. This implies that the number of bases before the last battery was exactly 10 (if $k_S \ge 1$). If $k_S=0$, $N=0$, $U_S=0$, which is consistent. If $k_S \ge 1$, then $N-(k_S-1)10 = 10$. $10k_S - 10k_S + 10 = 10$. This is true for any $k_S \ge 1$. So $N$ must be a multiple of 10, $N = 10, 20, 30, \dots$.

2. HAMMER (H): $n_H = 7$, $U_H = 3$. $N \equiv 3 \pmod{7}$. This means $N$ can be written as $N = 7k_H + 3$ for some integer $k_H \ge 0$. The final number of bases is $U_H=3$, which is less than $n_H=7$. This is consistent with the stopping rule. The number of bases before the last battery must be at least 7. $N - (k_H-1)7 \ge 7$ (if $k_H \ge 1$). $(7k_H + 3) - (k_H-1)7 \ge 7$ $7k_H + 3 - 7k_H + 7 \ge 7$ $10 \ge 7$. This is true for any $k_H \ge 1$. If $k_H = 0$, $N=3$. Then $U_H=3$, which is consistent ($3 < 7$). So, $N$ must be of the form $7k_H + 3$ where $k_H \ge 0$, i.e., $N = 3, 10, 17, 24, 31, \dots$.

3. BrahMos (B): $n_B = 3$, $U_B = 2$. $N \equiv 2 \pmod{3}$. This means $N$ can be written as $N = 3k_B + 2$ for some integer $k_B \ge 0$. The final number of bases is $U_B=2$, which is less than $n_B=3$. This is consistent with the stopping rule. The number of bases before the last battery must be at least 3. $N - (k_B-1)3 \ge 3$ (if $k_B \ge 1$). $(3k_B + 2) - (k_B-1)3 \ge 3$ $3k_B + 2 - 3k_B + 3 \ge 3$ $5 \ge 3$. This is true for any $k_B \ge 1$. If $k_B = 0$, $N=2$. Then $U_B=2$, which is consistent ($2 < 3$). So, $N$ must be of the form $3k_B + 2$ where $k_B \ge 0$, i.e., $N = 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, \dots$.

The initial number of Pakistani Air Bases $N$ must satisfy all three conditions simultaneously:

$N \equiv 0 \pmod{10}$

$N \equiv 3 \pmod{7}$

$N \equiv 2 \pmod{3}$

We need to find the minimum possible non-negative integer $N$ that satisfies these congruences. This is a problem that can be solved using the Chinese Remainder Theorem, but we can also find the solution by checking values or by building up the solution.

From $N \equiv 0 \pmod{10}$, $N$ must be a multiple of 10. Possible values for $N$ are $0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, \dots$.

Now let's check which of these values satisfy $N \equiv 3 \pmod{7}$:

$0 \div 7$ remainder is 0.

$10 \div 7$ remainder is 3. This satisfies the second condition.

Let's check if $N=10$ satisfies the third condition $N \equiv 2 \pmod{3}$:

$10 \div 3$ remainder is 1. This does not satisfy the third condition.

Let's continue checking multiples of 10:

$20 \div 7$ remainder is 6. (No)

$30 \div 7$ remainder is 2. (No)

$40 \div 7$ remainder is 5. (No)

$50 \div 7$ remainder is 1. (No)

$60 \div 7$ remainder is 4. (No)

$70 \div 7$ remainder is 0. (No)

$80 \div 7$ remainder is 3. This satisfies the second condition.

Let's check if $N=80$ satisfies the third condition $N \equiv 2 \pmod{3}$:

$80 = 3 \times 26 + 2$. $80 \div 3$ remainder is 2. This satisfies the third condition.

So $N=80$ satisfies all three conditions:

$80 \equiv 0 \pmod{10}$ (since $80 = 10 \times 8 + 0$)

$80 \equiv 3 \pmod{7}$ (since $80 = 7 \times 11 + 3$)

$80 \equiv 2 \pmod{3}$ (since $80 = 3 \times 26 + 2$)

Since we started checking from the smallest non-negative multiple of 10, the first value we found that satisfies all conditions is the minimum possible value of $N$.

So, the minimum possible number of Pakistani Air Bases is $x = 80$.

The question asks for the sum of digits of $x$.

The digits of $x=80$ are 8 and 0.

The sum of the digits is $8 + 0 = 8$.

The final answer is $\boxed{8}$.