The last two!

• The series of last digits 3 powers endings are 3,9,7,1,3,9.....

• Since the last digit is repeating for every fouth time, the 1997th power of 3 will end with 3.

• And the 1997th term will be the 500th term in the series ending with 3.

• The series of last two digits ending with 3 are 03,43,83,23,63,03,43.....

• Since the above series changing for every five terms,the 500th term ending with 3 will be ending with 63.

• Hence 63 is the last two digits of 3^1997.

The way to solve this problem is to find out what $3^{1997}$ is in mod 4 and mod 25, and then proceed to find it mod 100. Finding $3^{1997}$ mod 4 is simple: $3^{1997} \equiv (-1)^{1997} \equiv -1 \equiv 3 \pmod 4$ . Finding $3^{1997}$ mod 25 is slightly harder. The way I did was to note that $3^3 \equiv 2 \pmod {25}$ and $2^7 \equiv 3 \pmod {25}$ .

Then $3^{1997} = 3^{1995}*9 \equiv 2^{665}*9 \equiv 3^{95}*9 = 3^{93}*9*9 \equiv 2^{31}*81 = 2^{28}*8*81 \equiv 3^4*8*6 = 3^3*3*8*6 \equiv 2*3*8*6 \equiv 2*(-1)*6 \equiv -12 \equiv 13 \pmod {25}$ .

Thus $3^{1997} \equiv 3 \pmod 4$ and $3^{1997} \equiv 13 \pmod {25}$ . Mod 4 and mod 25 will cycle in 100, and the only number mod 100 that satisfies both these equations is $\boxed{63}$ . This can be found by repeatedly adding 25 to 13 until the resulting number is equivalent to 3 mod 4.

By Euler's theorem, we have $3^{\phi{(100)}}\equiv 1(mod 100)$ .On the other hand, we have the following

$3^{10}\equiv49(mod 100)$

and

$3^{30}\equiv(3^{10})^{3}\equiv49^{3}\equiv49(mod 100)$

and

$3^{7}\equiv49(mod 100)$

Thus,

$3^{1997}\equiv3^{1960+30+7}\equiv1\cdot49\cdot49\equiv63(mod 100)$

Therefore, the last two digits of $3^{1997}$ is $\boxed{63}$

$3^{1997}=3\times3^{1996}=3\times9^{998}$

last 2 digits for $9^{n}$ , n>=0:

01,09,81,29,61,49,41,69,21,89,01,09,...

we see that sequence repeats after 10 values;

998 mod 10 = 8;

$9^{998} = ...21, ...21\times3=...63$ , which is solution

Not as good as Datu Oen . But an alternative
$3^{15} \equiv 07~(mod~100)~~~~~~~~~~~~~~~~7^{4} \equiv 01~(mod~100)$
$3^{60}\equiv 01~(mod~100~~~~~~~~~~~~~~~~~~$ $3^{1997}\equiv 3^{17}~(mod~100)~$ $3^{15} *3^{2}\equiv 07*9~(mod~100) = 63$
$\boxed{63}$

3^1997=3(10-1)^998=3[(...)(100)+998(10)(-1)+1] (use binomial) =(...)(100)-29937=(...)(100)+63 =>so answer 63

3^1997 mod 100 => (3^4)^499 3 mod 100 => (81)^499 3 mod 100 => 21*3 mod 100 => 63

Note that $3^{1997} = 3^{1980} \times 3^{17}$ .

Since $3^{20} \equiv 1 \pmod {100}$ then $3^{1980} \equiv 1 \pmod {100}$ .

Thus we are left to finding the last 2 digits of $3^{17}$ which is 63 since $3^{17} = 3^{16} \times 3 \equiv (21) \times 3 \pmod{100}$ and $21 \times 3 = \boxed{63}$

3^1997 = 3^2 * 3^1997 = 9 * 87^285 = 9 * 7^57 = 9 * 43^19 = 87*49^3 = 49 * 87 = 4263 = 42[63] = 63. Answer : 63

At 3^n, the last two digits will repeat after n=20, it means that the last two digit of 3^1 = 3^21 ==> 03 3^2 = 3^22 ==> 09 3^3 = 3^23 ==> 27 . . . 3^1997 = 3^{ ( 20x99 ) + 17 } = 3^17 ==> 63

3^1=3 3^2=9 3^3=27 3^4=81 3^5=243 ...

The last digit (3 for 3^1, 9 for 3^2,....) repeats as 3,9,7,1.

Numbers ending in 3 will be of the form 3^(4n+1) Numbers ending in 9 will be of the form 3^(4n+2) Numbers ending in 7 will be in the form 3^(4n+3) Numbers ending in 1 will be in the form 3^(4n)

3^1997=3(1996+1)=3^1996 * 3^1

3^1996=3^(4*499)

Now numbers of the form 3^(4n) end in 1.

Also, the digits in the ten's place of this form follow the pattern (8,6,4,2,0) repeating after 5 numbers:

3^4=81 3^8=6561 3^12=531441 3^16=43046721 3^20=3486784401 3^24=282429536481

Then, 499=5*99 + 4 therefore , it's ten's digit will be 2 (similar to 3^16).

Therefore, 3^1996 will end in 21. Therefor 3^1997 will end in 3*21=63! ...

Therefor

use modulo :

the last two digits of 3^{1997}

= $3^{1997}$ mod 100

= ( $3^{4}$ )^499 . 3 mod 100

= $-19^{5}$ ^99. $-19^{4}$ . 3 mod 100

= $1^{99}$ . $-19^{4}$ . 3 mod 100

= 1. 21. 3 mod 100

= 63 mod 100

= 63

the last two digits of the number 7 show a cyclic pattern occurring every 4 powers, thus the given umber can be written as 27^665 X 3^2 . for the 27 part the cycle terminates at last two digits at 07 and the 3^2 multiplied to it gives us 63

Power Result Last 2 digits ===== ===== =========

1 3 03 2 9 09 3 27 27 4 81 81 5 243 43 6 729 29 7 87 (= 3 * 29) 8 61 (= 3 * 87 = 261, we discard the 2) 9 83 (= 3 * 61 = 183, we discard the 1) 10 49 (= 3 * 83 = 249, we discard the 2) 11 47 (= 3 * 49 = 147, we discard the 1) 12 41 (= 3 * 47 = 141, we discard the 1) 13 23 (= 3 * 41 = 123, we discard the 1) 14 69 (= 3 * 23 = 69) 15 07 (= 3 * 69 = 207, we discard the 2)

16 21 (= 3 * 7 = 21) 17 63 (= 3 * 21 = 63) 18 89 (= 3 * 63 = 189, we discard the 1) 19 67 (= 3 * 89 = 267, we discard the 2) 20 01 (= 3 * 67 = 201, we discard the 2) 21 03 (= 3* 1 = 3)