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Question 45832: What are the last two digits of:
(A)3 to the power of 1994
(B)7 to the power of 1994
(C)3 to the power of 1994 + 7 to the power of 1994
(D)7 to the power of 1994 - 3 to the power of 1994
Its something to do with congruences and modulo, but I dont understand what they mean.
Please show any working out
Thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! What are the last two digits of:
(A)3 to the power of 1994
Good problem.
I looked at the powers of 3 and found
a last digit pattern of 3,then 9, then 7, then 1.
I kept looking at powers of 3 till I got to
3^20 which has 01 as its last two digits.
After that point the preceding pattern of
last two digits repeats itself.
I divided 1994 by 20 and got a remainder of 14
So, as far as the last two digits is concerned
3^1994 = 3^(20*99+14)
3^14 has last two digits ...69) so I concluded
that 2^1994 has last two digits ...69
Hope this helps.
Cheers,
Stan H.
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