Math, asked by inshaf13, 9 months ago

GUESS!
147 - One digit is right but in the
wrong place
189 - One digit is right and in its
Place
964 - Two digits are correct but both
are in the wrong place
523 - All digits are wrong
286 - One digit is right but in the
wrong place
what are the three numbers ?
.​

Answers

Answered by Anonymous
278

Case 1)

147 - One digit is right but in the wrong place.”

Case 2)

189 - One digit is right and in it's (right) place.”

From Case 1) and Case 2) we can say that 1 is wrong. Because in both cases 1 is at the same place. So, 1 is eliminated or removed.

Case 3)

964 - Two digits are correct but both are in the wrong place.”

Case 4)

523 - All digits are wrong.”

As all digits are wrong. So, remove 5, 2 and 3 from all cases.

Case 5)

286 - One digit is right but in the wrong place.

From Case 2) Case 3) and Case 5) we can say that in Case 2) 9 is the correct digit and is placed in the right position.

So, we have __ __ 9.

Now from Case 1) and Case 3)

4 is in different positions. So, 4 is wrong.

Therefore, In Case 1) 7 is the correct digit but placed at wrong position.

From Case 2) Case 3) and Case 5)

8 is at the same position in Case 2) and 5) and correct in one case and wrong in another. So, 8 is eliminated.

Also, we can say that 6 is correct. From Case 5) it can't be at third position. Also, at third position we have 9. Also, it can't be at second position according to Case 3)

So, 6 is at the first position and 7 is at the second position.

Therefore, correct code is 6 7 9


BrainlyConqueror0901: fantastic bro : )
Anonymous: thanku :)
Answered by StarrySoul
136

Solution :

First of all exclude the numbers which aren't correct.

From the 4th data,We can conclude the digit 5, 2 and 3 are wrong.

Now,

• 147 - One digit is right but wrong placed

• 189 - One digit is right and placed correctly

• 964 - Two digits are correct but at wrong place

• 286 - One digit is right but in the wrong place

We've to find the three digits :

From 2nd,3rd and 5th data we can conclude that :

• Other set isn't having the digit 4

• Other set have either 6 or 9

From 1st and 2nd data we can conclude that :

• 9 is one of the three digits and it is correct placed

• 7 is the correct digit but wrong placed (4 and 1 is excluded)

Till Now We've :

→ _ 7 9 (We've got 7 from the first data and put it in the middle as according to given situation It can't be at end and we've got 9 from the second data and put it in the last as mentioned. )

In the 3rd data it was mentioned as two digits are correct. Since,9 is one of them and 4 isn't correct.

6 is the required digit and it can't be put in the middle as per given

Hence,The Three Digits are

6 7 9


BrainlyConqueror0901: woderful : )
StarrySoul: Thank you! :D
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