# If $a+b=10^{50}$ then show $10|a$

Suppose $a$ is a positive integer and another positive integer $b$ is obtained by jumbling up the digits of $a$. Suppose $a+b=10^{50}$. Then show that $10|a$.

I first looked at what happens if $a+b=100$. Letting $a=10x+y$, if $b\neq a$ then $b=10y+x$ implying $11(x+y)=100$, a contradiction. So $a=b$ meaning $a=50$ and hence $10|a$.

Then I looked at $a+b=10^3$. Here the situation is more complex, and I don't really see any easy pattern. Of course I also noted that if $10|a$ then $10|b$ also, so essentially by dividing both sides of $a+b=10^3$ by $10$ we can get $a^*+b^*=100$ and we are back in original situation, which has already been solved i.e. $a^*=50$ implying $a=500$ implying $10|a$ again.

So somehow I have to show that if $a+b=10^3$ holds, then $a=b$.

If I can tackle the smaller cases, I think I will be more capable of understanding what happens to $a+b=10^{50}$.

Oh, I also know that if $a$ and $b$ are numbers with same digits, then $9|a-b$. However, I don't know what else I can deduce.

• The case $a+b=10^3$ is more complex because it isn't true in that case. $455+545=1000$. As the answer shows, this is true for $a+b=10^{2n}$, however. However, the number must end in $0$ or $5$ in the odd case. – Thomas Andrews Jan 21 '17 at 14:06

Assume $a \not \equiv 0 \pmod {10}$. Note that $a$ must be smaller than $10^{50}$, yet bigger than $10^{49}$.

Let $$a=\sum_{i=0}^{49}a_{i}10^{i}$$ Then let $(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}, \dots, b_{49})$ be the digits of $b$ in order, and because of the definition of $b$ it follows that they are a rearrangement of $(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{49})$.

So $$10^{50}=\sum_{i=i}^{49}(a_{i}+b_{i})10^{i}$$

Since $a_{0} \neq 0$, we have that $a_{0}+b_{0}=10$. Because of carrying, we can use this to claim that $a_{1}+b_{1}=9$, and so on. Eventually, we note that for $n \ge i \ge 1$ we have $a_{i}+b_{i}=9$. So we have $$\sum_{i=i}^{49}(a_{i}+b_{i})=10 +9 +9+ \dots +9= 9 \times 49+10 =451$$ However, since $(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}, \dots, b_{49})$ is a rearrangement of $(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{49})$, we have $$451=\sum_{i=i}^{49}(a_{i}+b_{i})=2\sum_{i=0}^{49}a_{i}$$ A contradiction, since $\sum\limits_{i=0}^{49}a_{i}$ is a natural.

• Splendid! Thanks for the solution. – Landon Carter Jan 21 '17 at 13:49
• This proof is beautiful. Mind checking my proof maybe? Thanks~ – ryan w. Jan 21 '17 at 18:20

(I'd like to post an answer using more intuition and less fancy technique. The proof may look a bit weird at first, but I think it is valid. )

Step 1: show that the last digit of $a$ can only be $0$ or $5$.

Step 2: show that the last digit of $a$ cannot be $5$.

Now step 1:

Suppose the last digit of $a$ is $2$. Then the last digit of $b$ has to be $8$. Because $a$ is essentially a permutation of digits of $b$, so some digit of $a$ must be $8$, say its the $K^{th}$ digit of $a$. Then the $K^{th}$ of $b$ must be $1$ because $a$ and $b$ add up to $10^{50}$. By similar reasoning, $1$ must be some digit of $b$, say the $L^{th}$ digit of $b$. Then the $L^{th}$ digit of $a$ has to be $8$. In this way, number $1$ and $8$ reappears reiterately and number $2$ will never appear again. (You can try a few steps to understand the pairing system.) So $a$ is not a permutation of digits of $b$. A contradiction.

Using similar ways, we can show that the last digit of $a$ can only be $0$ or $5$.

Now step 2:

Suppose the last digit of $a$ is 5. Then the last digit of $b$ is 5. Then we have:

$$\overline{a5}+\overline{b5}=10^{50}$$$$\Rightarrow\overline{a0}+\overline{b0}=\underbrace{999...99}_{10^{50}-1}0$$$$\Rightarrow\overline{a}+\overline{b}=\underbrace{999...99}_{10^{50}-1}$$

We are going to prove that the last equation has no solution. Apply the same method in step 1. Suppose the last digit of $a$ is $3$. Then the last digit of $b$ is $6$. Then some digit of $a$ is $6$. The pattern in this pairing system is, $3$ and $6$ appear alternately. Note that $(10^{50}-1)$ is an odd number. So at last, there will be either one more $3$ or one more $6$. So $a$ is not a permutation of $b$. Applying similar reasonings, we have shown that the last digit of $a$ cannot be $1, 2, 3, 4, 5, 6, 7, 8, 9, 0$. So $a$ doesn't exist.

We have now proved that the last digit of $a$ can only be $0$.

• You're more than welcome to question the validity of the proof. – ryan w. Jan 21 '17 at 16:20
• Thanks ryan, I think your proof id correct. Unfortunately I have already accepted a best answer. – Landon Carter Jan 22 '17 at 2:55
• Haha, all good. I would accept his answer too if I were you. – ryan w. Jan 22 '17 at 3:03