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# help me!

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Find the last two digits of \$9^{8^7}\$. (By convention, exponent towers are evaluated from the top down, so \$9^{8^7} = 9^{(8^7)}\$.)

May 8, 2024

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The last two digits of a number depend only on the last two digits of the number itself. Note that the last two digits of the powers of 9 repeat in a cycle of 1: {1,9,81,65,49,21}.

We can start calculating the last two digits of some powers of 9:

91=9 (last two digits: 09)

92=81 (last two digits: 81)

93=729 (last two digits: 29)

94=6561 (last two digits: 61)

We see that the last two digits repeat in a cycle of 4: {09, 81, 29, 61}. Since 4 divides 7, the remainder upon dividing 7 by 4 is the same as the remainder upon dividing 87 by 4.

Because 8 is even, any power of 8 is also even, so 87 has a remainder of 0 when divided by 4. Therefore, the last two digits of 9^(8^7) are the same as the last two digits of 9^4, which is 61​.

May 9, 2024
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