MATH 135
Fall 2008
Assignment #6 Solutions
HandIn Problems
1. (a)
Answer
: YES
Since (

14)

22 =

36 = (

2)(18), then 18

(

14)

22, so

14
≡
22 (mod 18).
(b)
Answer
: YES
Since 2473

(

26) = 2499 = 147(17), then 17

2473

(

26), so 2473
≡ 
26 (mod 17).
(c)
Answer
: YES
Note that 8
2
≡
64
≡
4 (mod 12) and 8
3
≡
8
2
8
≡
4(8)
≡
32
≡
8 (mod 12). Thus, the
powers of 8 alternate being congruent to 8 and 4 modulo 12, with all even powers being
congruent to 4 modulo 12.
(d)
Answer
: 502
n
≡
0 (mod 4) if and only if 4

n
. Since 2008 = 4(502), then there are 502 multiples of 4
in this range.
(e)
Answer
: 2
Since 8
≡ 
1 (mod 3), then 8
243
≡
(

1)
243
≡ 
1 (mod 3).
Since the remainder is positive, then we rewrite this as 8
243
≡ 
1
≡
2 (mod 3).
Thus, 8
243
and 2 have the same remainder when divided by 3, which is 2.
(f)
Answer
: NO
Since 8
≡
1 (mod 7) and 13
≡ 
1 (mod 7), then
8
24
+ 13
12
≡
1
24
+ (

1)
12
≡
1 + 1
≡
2
(mod 7)
In other words, 8
24
+ 13
12
is not congruent to 0 modulo 7, so is not divisible by 7.
(g)
Answer
: 0
We look at small powers of 3 and of 5, until we ﬁnd one congruent to 1 or

1 modulo 13:
3
2
≡
9
(mod 13)
3
3
≡
27
≡
1
(mod 13)
and
5
2
≡
25
≡ 
1
(mod 13)
Thus, 3
47
= (3
3
)
15
3
2
≡
1
15
3
2
≡
9 (mod 13).
Also, 5
74
= (5
2
)
37
≡
(

1)
37
≡ 
1 (mod 13).
Next, we note that

9
≡
4 (mod 13), so we look at small powers of 4 modulo 13:
4
2
≡
16
≡
3
(mod 13)
4
3
≡
64
≡ 
1
(mod 13)
Thus, (

9)
10
≡
4
10
≡
(4
3
)
3
4
1
≡
(

1)
3
4
≡ 
4 (mod 13).
Therefore,
3
47
5
74
+ (

9)
10
≡
(9)(

1)

4
≡ 
13
≡
0
(mod 13)
Therefore, the remainder is 0.
(h)
Answer
: 11
For 14
≡ 
58 (mod
m
), we need
m

14

(

58) = 72, so
m
is a positive integer divisor
of 72 which is larger than 1.
Since 72 = 2
3
3
2
, then 72 has (3 + 1)(2 + 1) = 12 positive divisors, one of which is 1.
Thus, 72 has 11 positive divisors larger than 1.