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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
3 years, 8 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 \(1<a<\frac{10}{e}\). 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. Yong See Foo · 3 years, 8 months ago

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

Perfect Powers Abhimanyu Swami · 3 years, 8 months ago

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@Abhimanyu Swami Suppose \(d>ae+e\). Then \(1000e\leq e\cdot \overline{bcd}\leq 999e\), which is a contradiction. Yong See Foo · 3 years, 8 months ago

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You can prove it by setting e equal to 1 or e larger than 1 so there are less steps involved William Z · 3 years, 8 months ago

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Since \[1234*9\] is more than \[10000\] we find that e is less than 9. A case by case analysis might work here. Yash Talekar · 3 years, 8 months ago

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@Yash Talekar 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. Abhimanyu Swami · 3 years, 8 months ago

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