Solution.
This is Cantor’s other diagonal proof. We write the pairs of natural numbers
in a grid
(1
,
1)
(1
,
2)
(1
,
3)
(1
,
4)
· · ·
(2
,
1)
(2
,
2)
(2
,
3)
(2
,
4)
· · ·
(3
,
1)
(3
,
2)
(3
,
3)
(3
,
4)
· · ·
(4
,
1)
(4
,
2)
(4
,
3)
(4
,
4)
· · ·
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
and proceed down the “anti-diagonals” (which run from northeast to southwest) in turn,
giving the list
(1
,
1)
,
(1
,
2)
,
(2
,
1)
,
(1
,
3)
,
(2
,
2)
,
(3
,
1)
,
(1
,
4)
,
(2
,
3)
,
(3
,
2)
,
(4
,
1)
. . .
5(a).
Compute 3
43
(mod 8) using Euler’s theorem.
Solution.
First,
φ
(8) = 4.
(This is the number of numbers less than 4 and having
no factor in common with it, which are 1
,
3
,
5
,
7.)
Euler’s theorem states that
a
φ
(
n
)
≡
1
(mod
n
) whenever
a
and
n
have no common factor. Since 3 and 8 have no common factor,
this gives us 3
4
≡
1
(mod 8). (You could check this explicitly, since 3
4
= 81.) Thus
3
43
= (3
4
)
10
(3
3
)
≡
3
3
(mod 8)
and 3
3
= 27; the answer is the remainder when 27 is divided by 8, which is 3.
5(b).
Why can Fermat’s little theorem not be used in part (a)?
Solution.
Fermat’s little theorem only applies when the modulus is prime. 8 is not prime.
6.
Bank identification numbers consist of nine digits
n
1
, n
2
, . . . , n
9
, such that
7
n
1
+ 3
n
2
+ 9
n
3
+ 7
n
4
+ 3
n
5
+ 9
n
6
+ 7
n
7
+ 3
n
8
+ 9
n
9
≡
0
(mod 10)
.
Consider the bank identification number
211
X
72946
where
X
is a missing digit. What is
X
?
Solution.
Putting the claimed bank identification number into the formula, we get
7
·
2 + 3
·
1 + 9
·
1 + 7
·
x
+ 3
·
7 + 9
·
2 + 7
·
9 + 3
·
4 + 9
·
6
≡
0
(mod 10)
.
Doing the arithmetic gives 194 + 7
x
≡
0
(mod 10). So 7
x
≡
6
(mod 10); that is, 7
x
is
six more than a multiple of 10. A multiple of 7 which is six more than a multiple of 7 is
56 = 7
·
8, which can be found by trial and error; the answer is
x
= 8.
2