MAT 312/AMS 351, Summer 2011
Solutions to the Midterm I
Maximal grade: 100 points
1. (15 points) Find the last digit of the number
1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!
Answer:
3
.
Solution:
We have
1! = 1
,
2! = 2
,
3! = 6
,
4! = 24
.
For
n
≥
5 the number
n
! is divisible by 2 and by 5, so it is divisible by 10,
and its last digit is equal to 0. Therefore
1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! = 1 + 2 + 6 + 4 = 3
mod 10
.
2. (20 points) Compute 17
25
mod 9.
Answer:
17
25
= 8
mod 9.
Solutions:
a) Since 9 = 3
2
,
we have
ϕ
(9) = 3
2

3
1
= 6
.
By Euler’s theorem (we can use it since
g.c.d
(17
,
9) = 1) we have 17
6
= 1
mod 9
.
Therefore
17
25
= (17
6
)
4
·
17 = 1
4
·
8 = 8
mod 9
.
b) Remark that 17 = 8
mod 9, so
17
2
= 8
2
= 64 = 1
mod 9
.
1
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Therefore
17
25
= (17
2
)
12
·
17 = 1
12
·
8 = 8
mod 9
.
3. (10 points) Let
p
be a prime number. Find the maximal power of
p
that divides (
p
2
)! Prove that your answer is true for all prime
p
.
Answer:
p
+ 1.
Solution:
We have to count all multiples of
p
that appear in the defining
product for
(
p
2
)! = 1
·
2
·
. . .
·
p
2
,
that is, all multiples of
p
between 1 and
p
2
. These are
p,
2
p, . . . ,
(
p

1)
p
and
p
2
, the last one contributes to the second power of
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 Summer '08
 BESCHER
 Algebra, Number Theory, Prime number, Greatest common divisor, Coprime

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