Reversing Digits

Let N1=abcdN_{1}=\overline{abcd} and N2=dcbaN_{2}=\overline{dcba} be two 44 digits positive integers.

Let ee be a positive integer such that N1×eN_{1} \times e = N2N_{2}.

How many ordered sets of integers a,b,c,da,b,c,d and ee are there?

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

AA requestrequest : Please explain your steps a bit if you use Modular Arithmetic because I ain't much used to it.

Note by Abhimanyu Swami
5 years, 10 months ago

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3 votes

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Case-by-case analysis is not too bad. Consider e=1e=1. Then from the original statement we know that a=d,b=ca=d, b=c, which gives us 910=909\cdot 10=90 sets.

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

We then check for sets (a,d,e)(a,d,e) that satisfy the above inequality and aa is the last digit of dede. Note that we just need to check for a<10ea<\frac{10}{e}, which is not a lot, that is if we fix aa and ee, there are only 1212 valid pairs of (a,e)(a, e) just by 1<a<10e1<a<\frac{10}{e}. One gets (a,d,e)=(2,8,4),(1,9,9)(a,d,e)=(2,8,4), (1,9,9).

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

From the original statement, we arrive at 400b+40c+30=100c+10b400b+40c+30=100c+10b which simplifies to 390b+30=60c390b+30=60c. Therefore b<2b<2. Case-by-case analysis for b=(0,1)b=(0,1) gives us only (b,c)=(1,7)(b,c)=(1,7), which gives us the ordered set (a,b,c,d,e)=(2,1,7,8,4)(a,b,c,d,e)=(2,1,7,8,4).

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

Similarly, we have 900b+90c+80=100c+10b900b+90c+80=100c+10b which simplifies to 890b+80=10c890b+80=10c. Therefore b=0b=0, which gives c=8c=8, which gives us the ordered set (a,b,c,d,e)=(1,0,8,9,9)(a,b,c,d,e)=(1,0,8,9,9).

We then conclude there are 92 such sets.

Yong See Foo - 5 years, 10 months ago

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

Perfect Powers

Abhimanyu Swami - 5 years, 10 months ago

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Suppose d>ae+ed>ae+e. Then 1000eebcd999e1000e\leq e\cdot \overline{bcd}\leq 999e, which is a contradiction.

Yong See Foo - 5 years, 10 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 - 5 years, 10 months ago

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Since 123491234*9 is more than 1000010000 we find that e is less than 9. A case by case analysis might work here.

A Former Brilliant Member - 5 years, 10 months ago

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

Abhimanyu Swami - 5 years, 10 months ago

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