# 1983 AIME question

• Dec 25th 2006, 09:16 PM
Jenny20
1983 AIME question
Let a_n = 6^n +8^n.
Determine the remainder on dividing a_83 by 49.

Thank you very much.
• Dec 25th 2006, 10:46 PM
CaptainBlack
Quote:

Originally Posted by Jenny20
Let a_n = 6^n +8^n.
Determine the remainder on dividing a_83 by 49.

Thank you very much.

a_83=6^83+8^83

.......=(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]

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?)
• Dec 26th 2006, 12:47 AM
Jenny20
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....
• Dec 26th 2006, 01:01 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
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....

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
• Dec 26th 2006, 01:10 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
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....

Do exaxtly what I did but modulo 49, so:

6^83 + 8^83 = (7-1)^83 + (7+1)^83 ( mod 49)

and the rest works almost exactly as before

RonL

RonL
• Dec 26th 2006, 01:22 AM
Jenny20
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. :(
• Dec 26th 2006, 01:24 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
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=1162

and any numbers which are equal are congrunent to the each other modulo anything

RonL

RonL
• Dec 26th 2006, 01:37 AM
Jenny20
ic Thank you very much. :)