Finding the last two digit

Find the last two digits of

255575 9 832 . 2555759^{832}.


The answer is 81.

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2 solutions

255575 9 832 5 9 832 ( 5 9 5 ) 166 5 9 2 1 166 81 81 ( mod 100 ) 2555759^{832}\equiv59^{832}\equiv\left(59^5\right)^{166}\cdot59^2\equiv-1^{166}\cdot81\equiv 81\space(\text{mod 100})

What inspired you to realize that 5 9 5 1 ( m o d 100 ) ? 59^5 \equiv -1 \pmod{100}?

Eli Ross Staff - 4 years, 7 months ago

How about a number which is 2^12424536436234635415262456 ?

Kevin Verdhi - 4 years ago

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The last digit is 2.

Maja Drozd - 1 year, 9 months ago

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It won't be 2.

. . - 2 months, 2 weeks ago

Since 12424536436234635415262456 0 ( m o d 4 ) 12424536436234635415262456 \equiv 0 ( \mod 4 ) , so it is 6.

. . - 2 months, 2 weeks ago
. .
Mar 24, 2021

The last digit of 255575 9 832 2555759 ^ { 832 } is equal to 9 2 9 ^ { 2 } , so 1 1 .

The second digit is 8 8 . So, 81 \boxed { 81 } .

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