Question

Find the last two digits of3^1234

Answer 1

Answer:69

Step-by-step explanation:

3¹²³⁴

3¹²³²×3²

(3⁴)³⁰⁸ ×3²

81³⁰⁸×3²

81³⁰⁸ ×9

81³⁰⁸second last digit =8×8 ka last digit

last 2 digits =41

41×9=369

last 2 digits of 369=69 hence answer=69

hope you got it.

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Answer 2

Answer:

09

Step-by-step explanation:

Assuming you know basic Modular Arithmetic and Fermat's Little Theorem as your question is from elementary number-theory.

We have to find :-

$3^{1234} = ?(mod 100)$

By Fermat's Little Theorem ;

$3^{100} = 3(mod100) \\ 3^{99} = 1(mod 100)\\ 3^{1188} = 1(mod100)$

Now we multiply both the sides by 3^46 :-

$3^{1234} = 3^{46}(mod100)$

So basically we have now reduced the question from :

$3^{1234} = ?(mod 100)$

To This :

$3^{46} = ?(mod 100)$

From :

$3^{99} = 1(mod 100)$

We can say :

$3^{33} = 1(mod 100)\\ 3^{11} = 1(mod 100)$

Therefore :

$3^{44} = 1(mod 100)$

Multiplying both sides with 9 :

$3^{46} = 9(mod 100)$

Hence the answer is 09.