The EOs Have It
+ E E O O O O
The E positions in the cryptogram above indicate even digits and the O positions indicate odd digits. No digit is repeated. While there is not a unique way to fill in the cryptogram, the red O in the sum's result can only be one particular value. What is it?
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5 solutions
Simple surely:- Green O = 1, Orange O =3, Purple O =5 E = 2 Therefore Red O =9 9 IS CORRECT !!!!
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Tim, your answer results in both E's being 2. This violates the statement that "no digit is repeated".
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I made the same mistake he did. But the fact it it would have been more clear if the two E's had been different colors, since the odds were printed in different colors! That's what threw me off, thanks for clearing it up!
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@J.D. Waterhouse – Likewise - I mistook the two Es as being the same color.
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@Joe Harris – I changed one of the E colors to make it a further shade than the other blue was.
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@Jason Dyer – Perhaps it would actually be more clear if you removed all the colors.
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@John Young – That was how I originally had the problem, but the other editors preferred the colors, just to emphasize the digits being distinct. I don't think there's a way to make everyone perfectly happy; if nothing else, the instructions are clear about what to do.
Why he use the same symbols it makes confusing than the problem itself
this kind of problem, that tries to cheat the reader by causing them to overlook some subtle, insignificant and NONSENSICAL detail annoys me to the point of making me think of never again enter this f**ng web. I like solving problems, I do realize the importance of being aware of every detail, but I HATE being cheated. Would it have been so difficult to choose red color for E? In problems like this I can only see the desire of the author of cheating people. It is disgusting, to me...
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Or using regular letters for even and Capital letters for odd numbers, but DIFFERENT LETTERS. But, oh no, this would have being NOT trying to cheat people. Not funny, ain't it?!!! Meh!!
The E needs to be even, thus there must be a carry over of 2 from the sum of the odd digits on the right.
It can't have no carry over because then E = E contradicting the fact that the twos must be distinct digits
It can't have 1 carry over because if so, then E would be odd.
Thus, for 3 distinct odd digits to add up for a carry over of 2, then their sum must be ≥ 2 0
We choose the largest 3 odd digits which are 5 , 7 , 9 . Their sum lends up 21
Therefore O = 1
I have a request to Brilliant: can you guys change the color of the two E's? I really thought they were the same color and came up with the solution: 21 + 3 + 5 = 29. Of course, the odd digits could be in any order (23 + 1 + 5; 25 + 1 + 3; ...). In my thought process, I wanted to find three odd numbers that do not add up to more than 10, so there would not be carry and the tens digit would be the same. The only set of three distinct odd numbers that do not add up to more than 10 is {1, 3, 5} and they add up to 9, which was my answer.
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And I thought the two RED O's were the same......and as such, the problem was not solvable!!
E in the answer has to be two more than E in the sum. So the O's in the sum have to add up to be more than 20 to allow you to carry two over. So the only way is 9+7+5 = 21 This means that it is O in the answer = 1
It would help if the two Es were obviously different colours
Yep, the trick is to notice the carry.
The upper O have to be 5, 7, 9 so they add to 21 because any other choices for them add up to less than 20 and the lower E will not be even. So with these choices 2 Caries over and having the upper E even, the lower E will also be even.
The 2 Es must be different. Since addition is taking place then the carry from the addition of the 3 odds must be an even number greater than 0. Therefore 2,4,6... However, the greatest number 3 different odd digits can make is 5+7+9=21 (since these are the highest). Since it is 21 then the odd final digit must be 1.
When adding the 3 digits from the ones column, the carry number can be 0, 1, or 2.
The carry can't can't be 0; if it was, then the Es in the tens column would be the same, but it is given no digit is repeated.
The carry can't be 1, because an even number + 1 must be an odd digit, and the tens digit of the sum's result is even.
Therefore the carry must be 2. The only way to attain this from three distinct odd digits are 5 + 7 + 9 = 2 1 -- any smaller odd digits would result in a value less than 20 -- and so the ones place in the sum's result must be 1.