Difference of last two digits of 2019^(2019^(2019) ) is ?


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Question Number 41720 by Tawa1 last updated on 11/Aug/18

$Difference of last two digits of 2019^{2019^{2019}} is ?$
Commented by MJS last updated on 12/Aug/18
$=2019×2019×2019×... so we first need to know the sequence of the last 2 digits of 19^n : {19, 61, 59, 21, 99, 81, 39, 41, 79, 01, ...} the difference between these last 2 digits: {8, 5, 4, 1, 0, 7, 6, 3, 2, 1, ...} both have a period of 10 ⇒ we need to know the remainder of ((2019^(2019) )/(10)) the sequence of the remainders of ((2019^n )/(10)) {9, 1, 9, 1, ...} so the remainder is 9 if n is uneven and 1 if n is even 2019 is uneven ⇒ remainder is 9 ⇒ last two digits of 2019^(2019^(2019) ) =79 ⇒ difference is 2$
$= 2019 \times 2019 \times 2019 \times . . .$ $so we first need to know the sequence of the$ $last 2 digits of 19^{n} :$ ${19, 61, 59, 21, 99, 81, 39, 41, 79, 01, . . .}$ $the difference between these last 2 digits :$ ${8, 5, 4, 1, 0, 7, 6, 3, 2, 1, . . .}$ $both have a period of 10 \Rightarrow$ $we need to know the remainder of \frac{2019^{2019}}{10}$ $the sequence of the remainders of \frac{2019^{n}}{10}$ ${9, 1, 9, 1, . . .}$ $so the remainder is 9 if n is uneven and$ $1 if n is even$ $2019 is uneven \Rightarrow remainder is 9 \Rightarrow last two$ $digits of 2019^{2019^{2019}} = 79 \Rightarrow difference is 2$
Commented by Tawa1 last updated on 15/Aug/18

$God bless you sir$
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