A number theory problem by Rahil Sehgal

Number Theory Level 3

Find a 4-digit positive integer $a$ such that both $a^2$ and $a$ have the same last 4 digits.

The answer is 9376.

1 solution

Rahil Sehgal
Mar 1, 2017

As 76^{n} can have last two digits 76

376^{n} can have last three digits 376

9376^{n} can have last four digits 9376

Therefore a can only be 9376.

Thus, 10080-9376=704

how did you reach to the conclusion that only these numbers are possible like how you know 376 should be choose out of all 3 digit no .

sanyam goel - 4 years, 3 months ago

Mine was a kind of a long casework.

First conclude that $10000 \mid a(a - 1) \Rightarrow a = 625k , 625k + 1$ Then I checked all the cases( But a bit wisely).

Ankit Kumar Jain - 4 years, 2 months ago

@Jon Haussmann Sir , can you please share your solution?!!

Ankit Kumar Jain - 4 years, 2 months ago

These kinds of numbers are called automorphic numbers:

https://en.wikipedia.org/wiki/Automorphic_number

http://mathworld.wolfram.com/AutomorphicNumber.html

Rahil Sehgal has what is probably the easiest idea: build up the number one digit at a time. One class of automorphic numbers goes 5, 25, 625, 0625, 90625, etc. Another class goes 6, 76, 376, 9376, 09376, etc.

Jon Haussmann - 4 years, 2 months ago

@Jon Haussmann Thanks sir!

Ankit Kumar Jain - 4 years, 2 months ago

A number theory problem by Rahil Sehgal

The answer is 9376.

1 solution

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