Distribute numbers in a table

Relevant wikis:

• Modular arithmetic
• The Invariant Principle

The second row can contain numbers from 1 to 10 in any order: We cannot really use specific rules, we need to find a general one:

For example, we can try using the sum of all numbers in the second row: This sum will necessarily be $1+2+3+4+5+6+7+8+9+10=\frac{10\times(10+1)}{2}=55$ . The sum of all numbers in the first row will also be $55$ . Therefore, the sum of numbers in the first two lines will be $55+55=110$ .

Given the fact that the third line is simply an addition of the first two lines, the sum of all numbers in the third line will be $110$ , no matter what order we choose for the second line: Now we have a general, invariant rule! We can choose any order we want, the sum of all numbers in the third line will always be $110$ . Now let's see how we can use this to solve the problem:

If the unit digits of the numbers in the third row are all different, the sum of the numbers in the third row will be equal to: $S\equiv 45\equiv 5\pmod{10}$ (i.e. its last digit will be $5$ - see the associated wiki for more information). But we already know that the sum will be $110$ , whose last digit is $0$ : Contradiction!

Therefore the answer is $\boxed{No}$ .

In each of the first two rows, there are five even numbers $E$ and five odd numbers $O$ .

Let $0 \leq n \leq 5$ be how many even numbers in the second row are paired up with even numbers in the first row. Then we have

• $n$ combinations $E + E = E$ ;

• $5-n$ combinations $E + O = O$ ;

• $5-n$ combinations $O + E = O$ ;

• $n$ combinations $O + O = E$ .

Thus the bottom row will have $2n$ even numbers. However, if all the unit digits in the bottom row are different, this number should be equal to five. Clearly, this is impossible for an integer $n$ .

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Essentially my solution considers the problem modulo 2. The first and second rows have sums $\equiv 1$ modulo 2, therefore the bottom row must have sum $\equiv 0$ modulo 2, instead of 1. This is essentially the same argument as "No Info Given" gives in his/her solution, except that solution considers everything modulo 10.

Consider the general case: digits $0,\dots,n-1$ in the first two rows; is it possible for the sum row to contain all digits modulo $n$ ? If $n$ is odd, the answer is "yes". If top row and bottom row are equal, then the bottom row contains $0, 2, 4, \dots, 2n-2$ modulo $n$ , which turns out to be a permutation of the original digits. Also, in that case $n | 0 + 1 + \cdots + (n-1) = \tfrac12n(n-1)$ .

But if $n$ is an even number, we see that $n \not| \tfrac12n(n-1)$ , so that the bottom row cannot contain all the digits modulo $n$ .

If we start off by filling out the second row in order, we'll end up with 10 even numbers in the third row. We can get to any other possible ordering by switching around the numbers in the second row. For any single switch within the second row, there are three cases:

1) We switch two numbers that are both even. There is no change in the number even numbers in the third row.

2) We switch two numbers that are both odd. Yet again, no change in the number of even numbers in the third row.

3) We switch an even and an odd number. This will switch the parity of two numbers in the third row. This will result in gaining two evens, gaining a net of no evens, or losing two evens.

To have the all the units digits in the third row be different, we would require exactly 5 evens (0,2,4,6,8) and 5 odds (1,3,5,7,9). However, because we start with 10 evens and can only change the number of evens by 2, we will never be able to have exactly 5 evens.

The parity of any sum of integers is the parity of the sum of the unit digits of the integers. So we are looking for a solution when the sum of the third line is odd (sum of unit digits: 45). The sum of the third line is 2*55, which is even.

Kinda informal solution here, just to get the idea.

Summing the numbers with first row = second row will get the 3rd row as even numbers of all entries. That is, the third row has 10 even, 0 odd numbers.

If we have any 2 pairs of first row and second row, $(x_{1},y_{1}), (x_{2},y_{2})$ , we can always switch the numbers of the second row between the 2 pairs to $(x_{1},y_{2}), (x_{2},y_{1})$ . This operation is sufficient to generate the second row in any permutation.

In this way, 3 cases happen:

• $x_{1}+y_{1}, x_{2}+y_{2}$ are both odd/even, then $x_{1}+y_{2}, x_{2}+y_{1}$ either remains or both changes to odd/even.
• $x_{1}+y_{1}, x_{2}+y_{2}$ have the different parity, then $x_{1}+y_{2}, x_{2}+y_{1}$ either remains each of their own parity, or both changes their own parity.

In any cases, the third row's number of evens and odds can be changed such that 2 even numbers can be changed to 2 odd numbers or vice versa, or no changes at all in this operation.

Our desired case is that all numbers are different, which means there are 5 even, 5 odd numbers. As we have the possible start as 10 even, 0 odd, and the operations are bounded to increase/decrease by 2 (a.k.a. the parity of the numbers of evens/odds are invariant of this situation). Therefore, it's impossible to have such desired case.

If the answer was yes and since we have 10 boxes in the 3rd row, we must have a number in the third row whose units digit is 0( so 20) and a number whose units digit is 9(so 19) but we can't have both 19 and 20 in the third row because that would imply that there are two 10 in the 2nd row.

There is only 10 different digits to use in last row (0-9) and 10 boxes to fill so we need one digit in each box. But 3-rd row can't contain 0 and 1 (minimal sum is 2). So we are left with only 8 digits and right away we can say that it's impossible.