Mistakes give rise to Problems-9

Logic Level 4

In Roman Numerals, if you say that X L I V × X = C D X L XLIV\times X=CDXL , you're correct.

Otherwise, X L I V × X = C D X L \overline{XLIV}\times \overline{X}=\overline{CDXL} is a mistake and it's not necessarily correct.

For how many ordered six-tuplets ( C , D , I , L , V , X ) (C,D,I,L,V,X) of single digit positive integers ( x 0 ) (x\neq 0) 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

Satvik Golechha
Jul 29, 2014

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-

  1. 2864 × 2 = 5728 2864\times 2=5728

  2. 2814 × 2 = 5628 2814\times 2=5628

Thus, there are only 2 solutions.

That's good, now the problem is right. I am gonna include it in the set as soon as some moderator sees this and the red colored warning "This problem has been reported...." disappears. ;) By the way, in stead of "Sorry Aditya Raut" , you might like to say inspired from the set... and also say this problem is a part of that set, give link the way i do in the other problems of the set, so we'll make it a trademark together ! ;)

Aditya Raut - 6 years, 10 months ago

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@Aditya Raut Now the problem is problem-free, so you may like to add it in the set..

Satvik Golechha - 6 years, 10 months ago

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Ok, I made 3 problems for the set today, will post soon! I want you to tell me before you post a problem for the series. Hope you log in to the gmail to see messages. @Satvik Golechha

Aditya Raut - 6 years, 10 months ago

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@Aditya Raut @Aditya Raut See! This has become the officially toughest prob of the set!!

Satvik Golechha - 6 years, 10 months ago

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@Satvik Golechha Yeah, but there is one more problem of this set which is now 200 points, but at first it had got 315 315 points rating, you know why ? It had more than 1300 attempts, and has just 378 solvers. That had increased the rating so much that it became a 315 points question ! Then Calvin sir reduced it to 200 points :P

Aditya Raut - 6 years, 10 months ago

Wait... I though there could be duplications in the sixtuplets... Apparently this cost me two tries and I believe that you should have mentioned that.

Yannick Yao - 6 years, 10 months ago

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I've mentioned that the numbers are different.

Satvik Golechha - 6 years, 10 months ago

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Oops... My bad. Anyway it's also a good problem if identical digits are allowed.

Yannick Yao - 6 years, 10 months ago

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