IMO LONGLISTED PROBLEM 1970
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Prove that the two last digits of $9^{9^{9}}$ and $9^{9^{9^{9}}}$ in decimal representation are equal.
Re: IMO LONGLISTED PROBLEM 1970
Try with Euler. I am too lazy to post the solutions.
One one thing is neutral in the universe, that is $0$.
- Nadim Ul Abrar
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Re: IMO LONGLISTED PROBLEM 1970
The last two digits of $(9^{9})^{9}$= A is 09
and for $A^{9}$ that is 89
so only the last digits of A and A to the power 9 is equal
and for $A^{9}$ that is 89
so only the last digits of A and A to the power 9 is equal
$\frac{1}{0}$
Re: IMO LONGLISTED PROBLEM 1970
$9^{9^9}$ is not the same as $(9^9)^9=9^{9^2}$.Nadim Ul Abrar wrote:The last two digits of $(9^9)^9$= A is 09
"Everything should be made as simple as possible, but not simpler." - Albert Einstein
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla