IMO LONGLISTED PROBLEM 1970

MATHPRITOM · Unread post by **MATHPRITOM** » Sun May 01, 2011 1:23 pm

Prove that the two last digits of $9^{9^{9}}$ and $9^{9^{9^{9}}}$ in decimal representation are equal.

Unread post by **Masum** » Tue May 03, 2011 5:04 pm

Try with Euler. I am too lazy to post the solutions.

Nadim Ul Abrar · Unread post by **Nadim Ul Abrar** » Sat Jun 18, 2011 12:04 pm

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

Unread post by **nayel** » Sun Jun 19, 2011 6:25 pm

Nadim Ul Abrar wrote:The last two digits of $(9^9)^9$= A is 09

$9^{9^9}$ is not the same as $(9^9)^9=9^{9^2}$.

Nadim Ul Abrar · Unread post by **Nadim Ul Abrar** » Tue Jun 21, 2011 7:43 pm

Ops sorry ....

Nadim Ul Abrar · Unread post by **Nadim Ul Abrar** » Tue Jun 21, 2011 9:29 pm

The form of both number is $9^{40n+9}$......

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