# Reversing Digits

Let $$N_{1}=\overline{abcd}$$ and $$N_{2}=\overline{dcba}$$ be two $$4$$ digits positive integers.

Let $e$ be a positive integer such that $N_{1} \times e$ = $N_{2}$.

How many ordered sets of integers $a,b,c,d$ and $e$ are there?

It's not necessary that $a,b,c,d$ are distinct.

$A$ $request$ : Please explain your steps a bit if you use Modular Arithmetic because I ain't much used to it. Note by Abhimanyu Swami
6 years, 11 months ago

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Case-by-case analysis is not too bad. Consider $e=1$. Then from the original statement we know that $a=d, b=c$, which gives us $9\cdot 10=90$ sets.

Suppose now $e>1$. For the rest of the solution we have $(a+1)e>d>a$.

We then check for sets $(a,d,e)$ that satisfy the above inequality and $a$ is the last digit of $de$. Note that we just need to check for $a<\frac{10}{e}$, which is not a lot, that is if we fix $a$ and $e$, there are only $12$ valid pairs of $(a, e)$ just by $1. One gets $(a,d,e)=(2,8,4), (1,9,9)$.

Case 1: $(a,d,e)=(2,8,4)$

From the original statement, we arrive at $400b+40c+30=100c+10b$ which simplifies to $390b+30=60c$. Therefore $b<2$. Case-by-case analysis for $b=(0,1)$ gives us only $(b,c)=(1,7)$, which gives us the ordered set $(a,b,c,d,e)=(2,1,7,8,4)$.

Case 2: $(a,d,e)=(1,9,9)$

Similarly, we have $900b+90c+80=100c+10b$ which simplifies to $890b+80=10c$. Therefore $b=0$, which gives $c=8$, which gives us the ordered set $(a,b,c,d,e)=(1,0,8,9,9)$.

We then conclude there are 92 such sets.

- 6 years, 11 months ago

Can you explain, why is $(a+1)e>d$, By the way, nice solution. Now can you please solve this one?

- 6 years, 11 months ago

Suppose $d>ae+e$. Then $1000e\leq e\cdot \overline{bcd}\leq 999e$, which is a contradiction.

- 6 years, 11 months ago

You can prove it by setting e equal to 1 or e larger than 1 so there are less steps involved

- 6 years, 11 months ago

Since $1234*9$ is more than $10000$ we find that e is less than 9. A case by case analysis might work here.

- 6 years, 11 months ago

But won't that be too long? And remember I didn't say that $a,b,c,d$ are distinct. They may or may not be.

- 6 years, 11 months ago