18.781 Homework 1
Due: 9th February 2014
Q 1 (1.2(6). Prove that the product of three consecutive integers is divisible by 6; of four consecutive integers
by 24.
Proof. Let the integers be n, n + 1 an
18.781: SECOND PRACTICE HOUR EXAM #1
(for Hour Exam of March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply stat
18.781: SOLUTIONS TO SECOND PRACTICE
HOUR EXAM #1
(for hour exam of March 15, 203)
1. One way to do this is to write a = 20x + 7 and solve 20x + 7
8 (mod 43) using the Euclidean algorithm. We get x 2
18.781: PRACTICE HOUR EXAM #1
(for hour exam of March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply state the a
18.781: HOUR EXAM #1
(March 15, 2013)
This exam is closed book. Calculators are also not permitted. Do all four
problems. Be sure to show your work. Do not simply state the answer.
There are a total o
18.781: SOLUTIONS TO HOUR EXAM #1
(for hour exam of March 15, 2013)
1. (a) Write p2 p = (p 1)p, the base p expansion of p2 p. Since
p > 5, the base p expansion of 5p is just 5p. Hence by Lucas
theorem
18.781 Solutions to Problem Set 4, Part 2
1. (a) To nd a primitive root mod 23, we use trial and error. Since (23) = 22, for a to be a primitive
root we just need to check that a2 1 (mod 23) and a11 1
18.781: SOLUTIONS TO PROBLEM SET #2
(due February 19, 2013)
1.3(4) Let n =
ai 10i be the base 10 expansion of n, so 0 ai 9. The
sum of the digits of n is
ai . We have
n
ai =
ai (10i 1).
Now 10i 1 is d
18.781: SOLUTIONS TO PROBLEM SET #3
(due February 25, 2013)
2.1(17) By Wilsons theorem, we know that 70! + 1 0 (mod 71). Now
(70! + 1) (63! + 1) = 63! (70 69 64 1).
Since (63!, 71) = 1, we need to sho
18.781, Fall 2007 Problem Set 7 Due: FRIDAY, October 26
We are beginning a new section on algebraic number theory. Supporting discussion for this material can be found in Chapter 9 of NZM (though it i
18.781: SOLUTIONS TO PROBLEM SET #11
(due April 29, 2013)
6.2(4) Suppose to the contrary that a/b were such a rational number. Then
with the unique exception h = a , we have
k
b
ah
1
.
bk
bk
1
Hence i
18.781: SOLUTIONS TO PROBLEM SET #10
(due April 22, 2013)
5.4(4) Let 2a x, 2b y , and 2c z . Let u = x/2a , v = y/2b, and w = z/2c , so
u, v , w are odd. We may suppose by symmetry that a b c. If
x2 +
18.781: SOLUTIONS TO PROBLEM SET #8
(due April 8, 2013)
(A10) (1) This is a restatement of Theorem 2.20 on page 70.
(2) Let (m, n) = 1. Recall that if f (x) is a polynomial with integer
coecient and a
18.781: SOLUTIONS TO PROBLEM SET #7
(due April 1, 2013)
3.1(6) (a) Answer in back of book. To do the computation, simply com2
pute 12 , 22 , . . . , p1 and reduce modulo p. (See the rst part of
2
Prob
18.781: SOLUTIONS TO PROBLEM SET #6
(due March 18, 2013)
(A7) We have 8200 = (23 )200 = 2600 1 (mod 601) by Fermats theorem.
Hence ord601 (8) 200 < 600, so 8 is not a primitive root. In general, if
d|
18.781: SOLUTIONS TO PROBLEM SET #5
(due March 11, 2013)
2.7(2) Divide x7 x by 2x3 + 5x2 + 6x + 1 using the ordinary long division
algorithm for polynomials, but regarding the coecients modulo 7.
(You
18.781: SOLUTIONS TO PROBLEM SET #4
(due March 4, 2013)
2.3(4) By the Chinese remainder theorem, there will be a unique solution x
modulo 3 4 5 = 60. Trial and error shows that x 58 (mod 60). A bit
mo
18.781 Solutions to Problem Set 5
1. Note that 41 1 = 23 5. Start with a quadratic nonresidue mod 41, say, 3. Now b = 35 = 81 3 3
(mod 41), which has order exactly 8. (3)1 14 (mod 41).
Now we calculat
18.781 Solutions to Problem Set 2
k
+
1. Let m = n k. We want to show that the power of p dividing mk k = (m+k!)! is the number of carries
m!
when adding m to k in base p. Note that each time a carry
18.781 Homework 6
Due: 18th March 2014
Q 1 (2.8(2). Find a primitive root of 23.
Proof. We can check that 5 is a primitive root for the prime 23. (23) = 22, so 522 1(mod 23). But
52 = 25 1(mod 23) and
18.781 Practice Mid Term Exam #1
Esercizio 1. Find the multiplicative inverse of 22 modulo 135.
Solution
We perform the Euclidean algorithm for the GCD:
135 = 22 6 + 3
22 = 3 7 + 1
and going backwards
18.781 Introduction to Number Theory
Prof. A. De Sole
Mid Term Exam1
March 11, 2014
First and Last Name:
MIT ID number:
Problem #
Total Points
1
10
2
10
3
10
4
10
5
10
Total
50
Justify your answers!
1
18.781 Homework 3
Due: 25th February 2014
Q 1 (2.1(26). Show that the product of three consecutive integers is divisible by 504 if the middle one is a
cube.
Proof. 504 = 23 32 7. Let the three consecu
18.781 Introduction to Number Theory
Prof. A. De Sole
Mid Term Exam1
March 11, 2014
First and Last Name:
Alberto De Sole
MIT ID number:
1234567890
Problem #
Total Points
1
10
10
2
10
10
3
10
10
4
10
1
18.781 Homework 2
Due: 18th February 2014
Q 1 (1.3(3). Prove that a number is divisible by 2 if and only if the units digit is divisible by 2; that a
number is divisible by 4 if and only if the intege
18.781 Solutions to Problem Set 3
1. Its enough to solve the congruence mod 11 and mod 13, and then combine the solutions by Chinese
Remainder Theorem. Now x3 9x2 + 23x 15 factors as (x 1)(x 3)(x 5),