Soldiers after a battle

The units digit should either be 8 or 3 from condition 1. From condition 2, the number should be odd, hence units digit become 3. Now starting from 1200, we need numbers divisible by 11 and ending in 3. Check 1133. This does not give remainder 1 when divided by 7. Check 1023. This satisfies all conditions. Ans = 1023

first start with the given condition that it gives same remainder with 5 & 6,which means the number is of the form 30n+3.It is also given that it gives remainder 1 when divided by 7.Now seperating 30n+2 into 28n+2n+3 we can say that 2n+3 should be equal to 7n+1 where n is odd.this gives that the least number which fulfil the previous conditions simultaneously is 183 and the remaining numbers should be in a form 183+210n wich in turn is equal to 176+7+209n+n this will divisible by 11 only when 7+n=11k ,so for k=1,n=4.Now placing the value of n into our previous number we get-183+210*4=1023,which is fulfiling all the given condition.

11-7 = 4. So, to get 1 (mod 7), the smallest integer that is multiple of 4 is 8, that is 2 * 4 = 8 = 1 (mod 7).
So 2 * 11= 22 = 0 (mod 11) = 1 (mod 7)...<....>

22 = 4 (mod 6)...Next number to be added should be7 * 11 = 77 = 5 (mod 6)....
So 22+77 = 99 = 3 (mod 6).....
99 is good for 11, 7, 6.....

99 = 4 (mod 5)..<....>
Next number to be added should be 11 * 7 *6 = 462 = 2 (mod 5)..... So 99 + 462 = 571 = 4 + 2 (mod 5) = 1 (mod 5)

To get 3 (mod 5), we need to add 462 = 2 (mod 5). So 571 + 462 = 1023 = 3 (mod 5).
1023.

I made a small C++ program that checks every number between 11 and 1200 and if it is divisible by these 4 numbers, it will print it. It gave me the answer in 3 seconds ^_^

If we have bcd + acd + abd + abc, then in mod a, all but the first term become irrelevant in finding the value it is congruent to. The same applies to mod b, c, and d.

Thus, in mod 5, we have 6 7 11*x, where x is a number that makes the expression be congruent to 3 mod 5. x = 4 satisfies this constraint.

For mod 6, we have 5 7 11*3.

For mod 7, we have 5 6 11. (x = 1)

For mod 11, we have 5 6 7*11 (since it is congruent to 0 mod 11, all terms must be divisible by 11).

Thus, the sum of these is 5643. However, we can look at it in mod 5 6 7*11 (mod 2310). This is because the lcm of 5,6,7,11 is 2310 and adding/subtracting 2310 will not change the remainder when dividing by any of the original modulos.

Subtracting 2*2310 gives us 1023, which is the only answer in the range 0<answer<1200. Thus, our answer is 1023.

x mod 5=3 x mod 6=3 implies x mod 30=3.....(1) x mod 7=1.............(2) x mod 11=0...........(3) so from (1) 30a+3 is number 30a+3 mod 11=0 it implies 30(11m+1)+3 =330m+33 330m+33 mod 7=1 (from 2) 330(7y+3)+33 2310y+1023 is required value put y=0 for the above condition to get value less than 1200 soliders it gives 1023 as final answer

using Chinese remainder theorem

x = 3 * 462 * 3 + 3 * 385 * 1 + 1 * 330 * 1 + 0 * 210 * 1 = 5643,

N = 5 * 6 * 7 * 11 = 2310

now, 5643 - 2 * 2310 = 1023

hence, the answer is 1023

33, 143, 253, 363, 473, 583, 693, 803, 913, 1023, 1133.... all these positive real factors of 11 till 1200 and only 1023 satisfies the given conditions.

Let there be $x$ survivors. From the first 2 conditions, $x-3$ is a multiple of both 5 and 6, which is true if and only if $x-3$ is a multiple of 30. Therefore, $x$ can be represented as $30k+3$ for some integer k. We take this mod 11 to give that $10k+1$ is a multiple of 11. This is clearly true if $k=1$ and it is therefore true if and only if $k$ can be represented as $1+11r$ for some integer $r$ . Therefore, $x$ can be represented as $30(1+11r)+3=33+330r$ for some integer $r$ .

We also have that $x-1$ is divisible by 7, so we take $x-1=32+330r$ mod 7 to get $4+r \equiv 0$ (mod 7). The smallest positive $r$ that satisfies this is 3 (we can't have negative survivors, so $r$ is positive. Also, if there is a solution when $r$ is not minimal, there must also be a solution when $r$ is minimal because the only restriction on the number of survivors other than the fact that it is positive is an upper bound.)

Plugging in $r=3$ gives $x=33+330\times3=\boxed{1023}$ .

From first two conditions the number is 30x+3 From third condition we can say that 30x+2 is divisible by 7 now, 30x+2 = 28x + 2x +2 so, 2x+2 is divisible by 7 where minimum integer value of x is 6 meeting first three conditions, we get the number = 30 6+3 = 183 now, it is obvious that we have to maintain the value of x always incremented by 7, which means that for meeting the forth condition we set the number to 183+210y which is divisible by 11 183+210y = (176+209y)+(7+y) hence, y = 4 so, the number = 183+210 4 = 1023