Let a_n = 6^n +8^n.
Determine the remainder on dividing a_83 by 49.
Thank you very much.
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Let a_n = 6^n +8^n.
Determine the remainder on dividing a_83 by 49.
Thank you very much.
a_83=6^83+8^83Quote:
Originally Posted by Jenny20
.......=(7-1)^83 + (7+1)^83
Now expand the powers using the binomial theorem:
a_83=[(-1)^83 + 83 (-1)^82 7 + K_1 49] + [(1)^83 + 83 (1)^82 7 + K_2 49]
for some integers K_1 and K_2. So for some integer K_3
a_83=2 83 7 + K_3 49
.......= 1162 + K_3 49 = (23 49 + 35) + K_3 49
So the required remainder is 35.
RonL
(What does AIME stand for?)
Hi Captainblack,
Could you please solve this question by using the modulo 49? Thank you very much.
for example:
6^83 + 8^83 is congruent to x ( mod 49)
I don't know how to carry on the next working step after this....
Quote:
Originally Posted by Jenny20
a_83=6^83+8^83
.......=(7-1)^83 + (7+1)^83
Now expand the powers using the binomial theorem:
a_83=[(-1)^83 + 83 (-1)^82 7 + K_1 49] + [(1)^83 + 83 (1)^82 7 + K_2 49]
So
a_83=2 83 7 mod 49
.......= 1162 mod 49 = 35 mod 49
So the required remainder is 35.
RonL
Do exaxtly what I did but modulo 49, so:Quote:
Originally Posted by Jenny20
6^83 + 8^83 = (7-1)^83 + (7+1)^83 ( mod 49)
and the rest works almost exactly as before
RonL
RonL
a_83=2 83 7 mod 49
.......= 1162 mod 49
=======================
Hi Captainblack,
I don't get the above step = > 2837 mod 49 equal to 1162 mod 49.
In fact, I have completely forgotten the binomial thoerem. So feel confused with that part of your work. :(
Its a product: 2*83*7=1162Quote:
Originally Posted by Jenny20
and any numbers which are equal are congrunent to the each other modulo anything
RonL
RonL
ic Thank you very much. :)