Mistakes give rise to Problems-9
In Roman Numerals, if you say that , you're correct.
Otherwise, is a mistake and it's not necessarily correct.
For how many ordered six-tuplets of single digit positive integers is this mistake true?
Note: All letters must denote different numbers.
This problem is inspired from and is a part of the set:- Mistakes give Rise to Problems .
The answer is 2.
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1 solution
We have an alpha-numeric here. Note that since the product is a 4 digit number, 'x' must be 0, 1, 2, or 3. But 'x' can't be zero (given in the problem), and neither can it be 1, for then the product and the multiplier must be equal, which can't be possible because all letters are different numbers. Now, if x=3, then L can't be 0, since it is the first digit of the product, and 3*something will never end in 0 until that something becomes 0, which is again not possible because we can't have 2 letters for the same number. By similar bashing method, we have X=2. Similarly, we can find all the possible solutions, which are-
2 8 6 4 × 2 = 5 7 2 8
2 8 1 4 × 2 = 5 6 2 8
Thus, there are only 2 solutions.