Puzzle
- Thread starter Reverend
- Start date
Code:
786
786 - 876 = -90
-(9)0
Circle 9
Jumble all numbers and enter the all numbers apart from the circled one = 0
Jumble up = 0
Answer = 9
5891
5891- 1985 = 3096
(3)096
Circle 3
Jumble all numbers and enter the all numbers apart from the circled one = 96
Jumble up 69
Answer = 3Scratches head and remembers why I hated maths at school.
I didn't think much about it, but I guess it works somewhat like this:
When you subtract two numbers, for each digit basically only two types of thins happen: borrowing, or no borrowing. For each borrowing, 10 is added to the digit, and 1 is subtracted from another digit.
Now considering two digits (it's not in the game because it's too trivial). Since the two digits must be different, so there must be one borrowing, and the sum of the two resulting digits must be 10-1 = 9. Since the digits of the two numbers are the same, they are cancelled out after the subtraction. Therefore, it would be very easy to know the "circled" digit if another digit is given. Note that it's impossible to have no borrowing.
Now consider the real game, 3 or 4 digits. 3 digits first. There could be one borrowing, or two borrowing (three is, of course, impossible). For one borrowing, the sum is 9. For two borrowings, the sum is 18. How do you know how many borrowings happen? Since a digit ranges from 0 ~ 9, you can try to compute the digits by both two assumptions, and see if the resulting digit is in the range. However, if the sum of the two digits given is 9, the answer could be both 0 or 9. That's why the game prohibited you from choosing 0: the answer will be 9, not 0.
It's similar in the 4 digits case: the number of borrowings can be at most three (the sum is 27).
Now try to crack Tahir's case
Tahir gives 69, which sums to 15. 27 - 15 = 12 is too large for a single digit, but 18 - 15 = 3. That's the answer.
It's very late here (actually very "early" ) so I may make some mistakes, so feel free to correct me if I got it wrong.
When you subtract two numbers, for each digit basically only two types of thins happen: borrowing, or no borrowing. For each borrowing, 10 is added to the digit, and 1 is subtracted from another digit.
Now considering two digits (it's not in the game because it's too trivial). Since the two digits must be different, so there must be one borrowing, and the sum of the two resulting digits must be 10-1 = 9. Since the digits of the two numbers are the same, they are cancelled out after the subtraction. Therefore, it would be very easy to know the "circled" digit if another digit is given. Note that it's impossible to have no borrowing.
Now consider the real game, 3 or 4 digits. 3 digits first. There could be one borrowing, or two borrowing (three is, of course, impossible). For one borrowing, the sum is 9. For two borrowings, the sum is 18. How do you know how many borrowings happen? Since a digit ranges from 0 ~ 9, you can try to compute the digits by both two assumptions, and see if the resulting digit is in the range. However, if the sum of the two digits given is 9, the answer could be both 0 or 9. That's why the game prohibited you from choosing 0: the answer will be 9, not 0.
It's similar in the 4 digits case: the number of borrowings can be at most three (the sum is 27).
Now try to crack Tahir's case
Tahir gives 69, which sums to 15. 27 - 15 = 12 is too large for a single digit, but 18 - 15 = 3. That's the answer.It's very late here (actually very "early"
) so I may make some mistakes, so feel free to correct me if I got it wrong.
