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Find Remainders: 3^2002/26, 5^2002/26

```Date: 09/21/2002 at 22:10:23
From: Mr. Hugehy
Subject: Powers

(a)
Find the remainder obtained when 3^2002 is divided by 26
(b)
Find the remainder obtained when 5^2002 is divided by 26
(c)
Show that 3^2002 + 5^2002 is divisible by 26

```

```
Date: 09/22/2002 at 00:25:17
From: Doctor Greenie
Subject: Re: Powers

Hi, Mr. Hugehy -

Here is a link to a page in the Dr. Math archives where there is some

2^99 Divided by 99
http://mathforum.org/library/drmath/view/55599.html

If you use the technique described on that page, you will find that
the remainders when 3^n is divided by 26 for n=1 to n=10 are the
following:

3, 9, 1, 3, 9, 1, 3, 9, 1, 3

As you can see, the remainders when 3^n is divided by 26 repeat for
every 3rd power. Since the remainders of 3^n divided by 26 repeat
every 3rd power, and since 2002/3 = 667 remainder 1, the remainder
when 3^2002 is divided by 26 is the same as the remainder when 3^1 is
divided by 26.

the remainder when 3^2002 is divided by 26 is 3

And the remainders when 5^n is divided by 26 for n=1 to n=10 are the
following:

5, 25, 21, 1, 5, 25, 21, 1, 5, 25

Here, you can see that the remainders when 5^n is divided by 26 repeat
for every 4th power. Since the remainders of 5^n divided by 26 repeat
every 4th power, and since 2002/4 = 500 remainder 2, the remainder
when 5^2002 is divided by 26 is the same as the remainder when 5^2 is
divided by 26.

the remainder when 5^2002 is divided by 26 is 25

Once we have the answers to your first two questions, showing the
third should be easy.  If the sum

3^2002 + 5^2002

is to be divisible by 26, then the remainder when that sum is
divisible by 26 should be 0; this means the sum of the remainders
when 3^2002 and 5^2002 are divisible by 26 should be 0, or a multiple
of 26. However, we have found that the remainders when 3^2002 and
5^2002 are divided by 26 are 3 and 25; and 3+25=28 is not a multiple
of 26.

The problem here is that I think you have misquoted the problem.  I
have seen this same problem here at Dr. Math several times previously
in the last few months and every other time I have seen this problem,
the last part was to show that

3^2002 + 29*5^2002

is divisible by 26.

Since the remainders when 3^2002 and 5^2002 are divided by 26 are 3
and 25, respectively, then the remainder when

3^2002 + 29*5^2002

is divided by 26 is

3 + 29(25) = 3 + 725 = 728 = 28(26)

This remainder is a multiple of 26, and so the number

3^2002 + 29*5^2002

is divisible by 26.

I hope all this helps.  Write back if you have any further questions

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory
High School Number Theory

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