Question

Brainly challenge!!!!


??Who can solve this puzzle!!!!​

Answer 1

Answer:

The correct code is - $\boxed{0}\;\boxed{4}\;\boxed{2}$

Explanation:

This question is very Easy.

Let's do it,

Given Conditions :

682 – One number is correct and in the correct position.

645 – One number is correct but in the wrong position.

206 – Two numbers are correct but in the wrong positions.

738 – Nothing is correct.

780 – One number is correct but in the wrong position.

Now,

Let's see the first condition,

682 – One number is correct and in the correct position.

Here, It states that the one number is correct and in the correct position.

So, We can't understand anything from here.

Now, Let's see second condition,

645 – One number is correct but in the wrong position.

Hey! There my friend it states that One number is correct but in the wrong position.

Woah! :0 We can get some clue from here,

So, Let's combine first and second condition,

=> 6 8 2.

=> 6 4 5.

So, One number is correct and in the correct position.

Second - One states that One number is correct and in the correct position.

But we obeserve that, 6 is the common number in both, So, 6 can't be correct number. :P

It implies that,

We should cut 6 from both conditions. Now We have,

=> 8 2.

=> 4 5.

Now, Let's see the third condition,

206 – Two numbers are correct but in the wrong positions.  

Hey! Wait, here again six...So, No need to worries just cut that six :)..So, We have only 2 numbers in third condition. Hence, Both are correct numbers because third condition states that Two numbers are correct but in the wrong positions

. So, both are correct they will go to the lock but we cant find their correct places yet...No need to to worry. Let's find it out, :D

=> 2 0.

__________{ We have these as code. }

Now, Let's see the Forth condition,

738 – Nothing is correct.

Oh no, It is saying nothing correct So, We have to cut these number from every condition..

Let's see the fifth and last condition,

780 – One number is correct but in the wrong position.

So, We have to cut 7 and 8 from it because that numbers are wrong,

We have,

=> 0

=> 2 0

=> 8 2.

=> 4 5.

We will cut the 8 again and one zero is extra so, We will cut that too.

After cutting We have, :D

=> 2 0

=> 2.

=> 4 5.

Whole Explanation implies that zero is the first digit of the code,

Now, See third conditon, There is only 4 ,5 left :)

So, 4 is the third digit because Condtion say that one number is correct but wrongly placed.

Now, Let's see second condition,

There is,

=> 2.

So,

Condition states that,

One states that One number is correct but in the wrong position.

Here, 1 can't be the code digit because condition say that number is wrongly placed.

Hence, 2 is the second code digit.

Our code is :

$\boxed{0}\;\boxed{4}\;\boxed{2}$