Find the last $2$ digits of $3^{2024}$ using binomial expansion

Question

Find the last two digits of $$3^{2024}$$

I can easily do this question (see below) but is there a way of doing it by binomial expansion? I tried to expand the expression to a few terms but in vain.

Any help is greatly appreciated.

EDIT

By seeing some comments, I am compelled to ask if this question can really be solved using binomial$?$

EDIT My solution without binomial expansion: $$3^{15}\equiv 7\pmod{100}\implies$$ $$3^{2010}\equiv 7^{134}\equiv7^2 \pmod{100}\implies$$ $$3^{2024}\equiv 7^2\cdot3^{14}\equiv 7\cdot3^9\equiv 3^4 \pmod{100}$$ Hence the answer is $3^4$ i.e. $81$

Binomial expansion $(a+b)^{2024}$ for which $a,b$? And how did you use the Chinese remainder therorem? — Anne Bauval, Oct 07 '22 at 12:59

score 1 · Accepted Answer · edited Oct 07 '22 at 13:24

1

(Per preferred_anon's comment.)

Note that

$$3^{2024} = 9^{1012} $$ $$= (10-1)^{1012} = 10^{1012} - {1012\choose 1}10^{1011}+\cdots+{1012\choose 1010}10^2 -{1012\choose 1011}10 +1 $$

$\equiv -10120+1 \equiv -20 +1 \equiv 81 \pmod{100}.$

edited Oct 07 '22 at 13:24

Anne Bauval

34,650

answered Oct 07 '22 at 13:22

B. Goddard

1

5

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 07 '22 at 13:31

Find the last $2$ digits of $3^{2024}$ using binomial expansion

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