I can't use all 10 digits here!
+ 2 □ □ □ 5 □ □ □ 5 □ □ □ 5
The above is a long addition with some numbers hidden in the boxes. Each box represents a distinct digit from 0 to 9.
Fill up all the boxes and find the digit not used.
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7 solutions
The only shortcoming I see is that you don't show that a solution actually exists. Though I suppose one could argue that the problem is predicated on the existence of a solution.
I did the same thing. I thought about actually finding the digits that go in the boxes… But then I decided, nah, I didn't want to find a pencil and paper.
There is the one answer of "1." There are 27 possible sets of additions by varying the orders of the 7, 8, 9 and the 4, 5, 6 and the 0, 2, 3, in their respective columns.
i solved it by creating a matlab code :v
650 + 951 + 954 = 2555
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You haven't used distinct digits. Moreover, what you've written doesn't lend itself to a single digit answer.
997+997+601=2555 so i don't get it, you can do it with a '1' in one of the numbers
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each number 0-9 is used only once
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Damn, I made the same mistake, didn't notice the "distinct" word.
Notice that the maximum sum of a particular column is 9 + 8 + 7 = 2 4 . The leftmost column will need to be 9 , 8 , 7 because the maximum sum is 2 4 , and we need 2 5 , therefore, it's reasonable to assume that the leftmost column needs to achieve maximum value:
+ 2 9 8 7 5 □ □ □ 5 □ □ □ 5
Now, observe that we need a 1 carry-over from the middle column in order to make 2 5 from the leftmost column. There are several ways we can construct to obtain 1 5 . However, the most straightforward way is:
+ 2 9 8 7 5 6 5 4 5 □ □ □ 5
Now the rightmost column has these numbers left: 0 , 1 , 2 , 3
The only possible way to obtain 5 from these numbers is:
+ 2 9 8 7 5 6 5 4 5 3 2 0 5
Therefore, 1 will be left out, unused.
This was the way I solved it as well. Perhaps given the confusion this seems to have caused some solvers such questions in future could use the 'belt and braces' defence of not only stating that each box is filled by a distinct single digit between 0 and 9 but also actually stating rather than leaving it to the solver that each digit can only be used 1 (I personally did not have difficulties with this - to me it was a natural assumption, but this solution is not the only one to feature comments from people who misread the question in this way).
This is the approach I took as well.
However, while there are several ways of constructing 15 from the sum of three unique digits from 0 to 9; Once the 9,8,7 digits have been established as necessary in the first column, I can only see one way of constructing 15 with three of the remaining unique digits (even when considering a carry of 1).
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15 could also come from 6+5+3 but that requires a carry of one. That would mean that the remaining 3 digits would have to also add to 15. That isn't possible with 4, 2, 1, and 0. Thus the second column must be 6+5+4 and the last column 2 + 3 + 0
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Precisely.
There is an argument for the wording being an issue in this solution. Though, I appreciate that English might not be Vu Vincent's first language.
By stating that there are several way of constructing 15 in the 2nd column, this solution could be alluding to the permutations of the 6,5,4 digits. But then, it claims that there is only one possible way to obtain 5 in the 3rd column, which implies that permutations of digits within each column are being disregarded.
This inconsistency either leaves the wording about the 2nd column as wrong, or the wording about the 3rd column as wrong.
- If we disregard permutations of digits, there aren't several ways of constructing 15 in the 2nd column (as you confirmed).
- If we do consider permutations of digits, there is more than one way of obtaining 5 in the last column.
But I might just be being unnecessarily picky about the language being used in this solution.
The same approach mine as well. I Just note down the all 10 digits from backward 9 , 8 , 7 , 6 , 5 , 4 , 3 , 2 , 1 , 0
I see how you did it. My issue is not with your answer but the wording of the question. "A distinct digit in each box" does not preclude the multiple use of digits. It simply means no other digits than the ones noted to be available.
I think you meant 1 will be left out, unused
Let's be lazy.
Since 1 0 0 ≡ 1 0 ≡ 1 modulo 9, we know that − 1 ≡ 2 5 5 0 ≡ ( 1 + 2 + ⋯ + 9 ) − c = 4 5 − c ≡ − c mod 9 . It follows that c = 1 .
Or a little less lazy.
The carry from each column is at most two. Thus the sum of the leftmost column lies between 23 and 25:
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25 is impossible, because 9 + 8 + 7 = 2 4 is the highest possible value.
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23 would be possible with 9 + 8 + 6 = 2 3 , but that leaves nothing more than 7 + 5 + 4 = 1 6 for the middle column, making it impossible for it to carry two to the first column.
So we settle on a total of 24 for the first column, with digits 9 + 8 + 7 . Next, the second column plus carry must result in 15. This can be done in two ways; we already ruled out a carry of two from the rightmost column.
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14 in the middle column requires a total of 15 in the last column. The total of the columns would be 2 4 + 1 4 + 1 5 = 5 3 . But the sum of all digits 0 through 9 is only 45. Therefore this is not possible.
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We must have 15 in the middle column and 5 in the last column. The total of the columns is 2 4 + 1 5 + 5 = 4 4 , one short of 45. Therefore the missing digit is 1 .
Final details: 5 = 0 + 2 + 3 is the only possibility for the right column, leaving us with 1 5 = 6 + 5 + 4 for the middle column. The precise location does not matter, as the numbers in the columns can be permuted at will. Thus there are 3 ! 3 = 2 1 6 possible ways to fill out the grid, and 3 6 different set of three addends.
Could you expand on the 'lazy' solution ? From quick readings I found you're using modular arithmetic (which I had never heard of and I'm now quite curious about) to find the remind(er)ing digit but I struggle to understand how you applied it to the problem - and I'm confused by the comparison between your answer (in which you leave 0 out and use mod10) and Khang Nguyen Thanh's answer (in which he uses mod9 and keep the 0). No need for a full detailed explanation, as I will read a bit more about it when I have more time, but a few hints would be appreciated. Thanks in advance.
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My bad-- I used arithmetic modulo 9. Sorry for the typo!
The thing I don't understand in the solutions above is the middle right column (15)...don't you have to carry 2 from 24 to the left? That results in 17, not 15 in the next column...
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No, I worked from left to right, not right to left.
This is a logic problem. What I do know right now is that the largest numbers are used, because the result is a large number. Even if we were to use a small number 9x it would not add up. So start up and work down.
Working from highest to lowest, I know that none of the larger numbers can be eliminated, and we aren't given the choice to eliminate any of them. Further, I know that if we are working from highest to lowest, we do so without skipping any numbers. Logically, skipping numbers also makes no sense, since we don't know which numbers to skip. So we know that one of the lowest numbers is to be be eliminated, and we logically know that that number is not 0, because 0 adds nothing to the long addition. It doesn't help it or harm it. So 0 is eliminated, and it was eliminated for us. Logically, there is no reason why we cannot use all of the numbers in some combination, but by working from highest to lowest and knowing that one number has to be eliminated, we know that the one number that has to be eliminated is the last one. It would be the number left unused after we've worked downward.
The answer is 1.
I solved it from left to right. The left column has to sum 25, a high number to sum only with one digit numbers. Remember the existence of the carry, i.e. the overflow of the middle column. The carry cannot be 2 because if we have a carry of 2, it means that we need two sums of 25, what is not possible with the available numbers, and cannot be 0 because the higher numbers available (9 + 8 + 7) sum only 24, so we need the carry of 1. This means that the middle column has to sum 15. Then we two possibilities in the carry of the middle: 0 or 1. If the carry is 1, we need to sum 14 in the middle column and 15 on the right column, what is not possible with the numbers there are still available, so we can't have a carry, and sum 15 at all. The only possibility for this is 6 + 5 + 4. It left us with the numbers 0, 1, 2 and 3 to sum 5 at the right column, summing 6. We just need to exclude the number 1 and voilà, we have our answer.
Excellent deduction. Solving it from left to right really is more approachable than going from right to left.
1 is left out. I did it by digital roots (sum of the digits, until you have just one digit). 2+5+5+5=17 >>> 1+7=8 (digital root). 1+2+3+4+5+6+7+8+9=45 >>>4+5=9 (digital root). That is 1 too high, so the digit that needs to be left out is 1. You don't need to spend time calculating where any of those digits go.
How does this work? It looks like some sort of sorcery to me.
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He meant that he's applying the divisibility rule of 3, but that's not obvious in his solution. Plus, he didn't prove that such a configuration exists.
This is a logic problem. What I do know right now is that the largest numbers are used, because the result is a large number. Even if we were to use a small number 9x it would not add up. So start up and work down.
Working from highest to lowest, I know that none of the larger numbers can be eliminated, and we aren't given the choice to eliminate any of them. Further, I know that if we are working from highest to lowest, we do so without skipping any numbers. Logically, skipping numbers also makes no sense, since we don't know which numbers to skip. So we know that one of the lowest numbers is to be be eliminated, and we logically know that that number is not 0, because 0 adds nothing to the long addition. It doesn't help it or harm it. So 0 is eliminated, and it was eliminated for us. Logically, there is no reason why we cannot use all of the numbers in some combination, but by working from highest to lowest and knowing that one number has to be eliminated, we know that the one number that has to be eliminated is the last one. It would be the number left unused after we've worked downward.
There are so many other sums such as 753+860+942, how are you sure that none of them uses 1?
I used a lazy heuristic: You carry the one in long addition. Therefore 1 is not used in the sum. Simples. Tsk
Where do you carry the one in long addition, exactly?
I solved without filling up any box.
Because a number and sum of its digits has the same remainder when divide by 9 and 2 5 5 5 ≡ 8 ( m o d 9 ) , sum of all used digits leaves a remainder of 8 on division by 9 .
Combining with 0 + 1 + 2 + … + 9 = 4 5 ≡ 0 ( m o d 9 ) , we get the digit not used must leaves a remainder of 1 on division by 9 .
So, the digit not used must be digit 1 .