Form 10^10 With The Same Digits

Clearly $C$ and $D$ must be 10 digit numbers. Let $C = \overline{c_1 c_2 \ldots c_{10}}$ and $D = \overline{d_1 d_2 \ldots d_{10}}$ . Observe that $d_{10} + c_{10} = 0$ or 10.

Suppose that $d_{10} + c_{10} = 10$ . Then, this implies that $d_i + c_i = 9$ for $i = 1$ to 9. This gives

$2\sum d_i = \sum (d_i + c_i) = 9 \times 9 + 10 = 91.$

This is a contradiction as the LHS is even, while the RHS is odd. Thus, our original assumption that $c_{10}+d_{10}=10$ is wrong.

Hence, we must have $d_{10} + c_{10} = 0$ , which gives us $c_{10} = d_{10} = 0$ .

Okay, this one is tricky. Consider that the problem is symmetric for $C$ and $D$ ; that is, one of the numbers can be rearranged to form the other. So the last digit of them both has to be the same, since we're working with modulo 10 here (Here I'm stating that $C == D (mod 10)$ , otherwise we'd have more than one solution.). Now, the only way that two numbers which have the same last digit can be summed and end with a $0$ are either both ending in $0$ or both ending in $5$ .

If the numbers end in $0$ , then everyone divides $10$ and we can all be happy about it. If the numbers end in $5$ , however, there's a carry of $1$ to the tens position while performing the summation. Now, the only way to guarantee that the tens position will be a $0$ is by having the tens position of $C$ and $D$ having a sum of 9 (without taking the carry into factor). Notice that this will cause a ripple effect throughout the sum; every position will have a carry of $1$ from the previous calculation. Therefore, each and every pair on every position must have a sum equal to $9$ to carry on.

This is where it gets interesting. Let $C_{i}$ be the i-th digit of $C$ , and $D_{i}$ be the i-th digit of $D$ . If there's a pair of digits $(C_{i}, D_{i})$ satisfies the condition set on the last paragraph that $C_{i} + D_{i} = 9$ , then there must be another pair of numbers $(C_{j}, D_{j}), i \not= j$ in the sum such that they satisfy $C_{i} = D_{j}$ and $C_{j} = D_{i}$ . This is true because $C$ is a permutation of $D$ , and thus each digit of $C$ is locked to $9$ minus itself, which by the pigeonhole principle can make us reach this conclusion automatically.

Now, keep in mind that the only ordered pairs of numbers that satisfy $C_{i} + D_{i} = 9$ are: $(0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)$ . (As mentioned in the end of the last paragraph.) So, if we have a pair $(C_{i}, D_{i})$ that is equal to, let's say, $(2, 7)$ , there must then be another pair $(C_{j}, D_{j})$ which is the pair $(7, 2)$ .

Finally, notice that $10^{10}$ is a 11-digit number, so it can only be the sum of two numbers with at most 10 digits. If we chose the ones place of $C$ and $D$ as being equal to 5, there will only be 9 pairs of $(C_{i}, D_{i})$ left to distribute along the number. Problem: For each pair, its "opposite" must also be contained in both numbers. But since we only have 9 pairs left, there will be a pair left out, since each pair is unique and cannot be replaced with itself. Thus, there is no number ending in 5 which can be rearranged and summed with itself to form $10^{10}$ . Thus, the remainder of the division must be 0.

The simplest example of solution I could find was $D = 5450000000$ and $C = 4550000000$ , which illustrates the result. Of course, there are others.

(Note: The argument of the "carry ripple" can be used to give a stronger proof that both $C$ and $D$ must end with the same digit, but for simplicity's sake I decided to not write it down because it would be a waste of time.)

We can get a sum of 9 by (1+8,2+7,3+6,4+5) and for each combination among the four there should be reverse of it so that the all the digits are same in C and D . So, the first 8/6/4/2 digits of C and D can be any combination from the four combinations along with reverse of it (8+1,2+7,3+6,5+4) as number of combinations should always be even.

Now the last 2/4/6/8 digits will have to make a sum of 100/10000/100000/100000000 respectively in such a way that the digits are same in C and D leaving a carry of 1 . This can happen only when the last 2/4/6/8 digits are 50/5000/500000/50000000 respectively .

I used a method that was not very mathematical.

If two numbers have same digits and add up to a power of 10, they can be fractions of 7, to a power of 10 (For example: 4285710000 + 5714280000 = 10000000000, or (3/7) 10^10 + (4/7) 10^10 = 10^10)

Thus, the first 6 digits are non-zero and belong to both D and C. So, the last 4 digits of D and C will be 0. Thus, the remainder when divided by 10 is 0.

If sum of two numbers having same digit is in order of 100 then then they are 50\500\5000 so when we divide remainder is always 0.

Check this instinct ..... since both the numbers sum up to 10^10 so the units digits must have (1,9), 2,8) and so on but as their number of digits are same so so clearly all other digits except 0,0 are eliminated or else (as I do not knew the trick I found it mechanically ) the two numbers will have separate digits so the answer follows