This preview shows pages 1–2. Sign up to view the full content.
MATH 135, Fall 2010
Solution of Assignment #7
Problem 1
.
Suppose that
a
= (
r
n
r
n

1
···
r
2
r
1
r
0
)
10
3
. This notation means that
a
= 10
3
n
r
n
+ 10
3(
n

1)
r
n

1
+
···
+ 10
3
r
1
+
r
0
where each of
r
n
,r
n

1
,...,r
1
,r
0
is a 3digit number between 0 and 999, inclusive. Prove that 13

a
if and only if 13

r
0

r
1
+
r
2

r
3
+
···
+ (

1)
n
r
n
.
Solution.
Considering
a
modulo 13, we obtain Note that 1000
≡ 
1 (mod 13).
a
≡
10
3
n
r
n
+ 10
3(
n

1)
r
n

1
+
···
+ 10
3
r
1
+
r
0
(mod 13)
≡
(

1)
n
r
n
+ (

1)
n

1
r
n

1
+
···
+ (

1)
r
1
+
r
0
(mod 13)
(since 1000
≡ 
1 (mod 13))
≡
r
0

r
1
+
···
+ (

1)
n

1
r
n

1
+ (

1)
n
r
n
(mod 13)
Therefore,
a
≡
r
0

r
1
+
···
+(

1)
n

1
+(

1)
n
(mod 13), and so 13

a
if and only if
a
≡
0 (mod 13)
if and only if
r
0

r
1
+
···
+(

1)
n

1
+(

1)
n
≡
0 (mod 13) if and only if 13

r
0

r
1
+
···
+(

1)
n

1
+
(

1)
n
, as required.
Problem 2
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 12/10/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Math

Click to edit the document details