Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

8 Problems with Solutions on Introduction to Proofs Via Number Theory | MATH 302, Exams of Mathematics

Material Type: Exam; Class: Introduction to Proofs Via Number Theory; Subject: Mathematics; University: Western Washington University; Term: Fall 2007;

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

koofers-user-bl4
koofers-user-bl4 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MATH 302 Test 2 Solutions
November 2, 2007 Name
Put your answers in the space provided. Show your reasoning. Calculators are to be used when appro-
priate. The maximum score on the test is 30 points.
1. 4 points Without using your calculator find the remainder when 20072007 is divided by 13.
2007 707 57 5 (mod 13) 2007 807 87 3 (mod 12)
20072007 512·s+3 512s·531s·25 ·5 58 (mod 13)
So the remainder is 8
2. 4 points Find the two smallest positive solutions to
2x8 (mod 4) 3x9 (mod 5) 4x10 (mod 6)
Our system is equivalent to
x0 (mod 2) x3 (mod 5) x1 (mod 3)
So x= 2a2a38 (mod 5) a4 (mod 5) a= 4 + 5b
x= 2(4 + 5b) = 8 + 10b8 + 10b1 (mod 3) b 72 (mod 3)
b= 2 + 3c
x= 8 + 10(2 + 3c) = 28 + 30cThe two smallest positive solutions are 28 and 58
3. 3 points Theorem: If a|cand b|cand if (a, b) = 1, then ab|c
Proof:
1. c=as,c=bt definition of division
2. ax +by = 1 for some x, y Since (a, b) = 1
3. cax +cby =cMultiply both sides by c
4. btax +asby =csubstitute from 1.
5. ab(tx +sy) = cROF, factor out ab
6. ab|cdefinition of division
In the boxes fill in a short reason for that step. An acceptable reason is ROA (rules of algebra).
4. 4 points Characterize those positive integers nfor which φ(n)
2is odd. Clearly indicate your answer.
φ(4)
2= 1. If n=pαfor some odd prime p, then φ(n) = n 11
p!=n p1
p!.
So if φ(n)
2is odd, 4does not divide p1or p3 (mod 4)
φ(2) = 1. So if (n, 2) = 1, then φ(2n) = φ(n)
n= 4 or if n=pαor n= 2pαwith pprime and p3 (mod 4), then φ(n)
2is odd.
pf2

Partial preview of the text

Download 8 Problems with Solutions on Introduction to Proofs Via Number Theory | MATH 302 and more Exams Mathematics in PDF only on Docsity!

MATH 302

Test 2 Solutions November 2, 2007 Name

Put your answers in the space provided. Show your reasoning. Calculators are to be used when appro- priate. The maximum score on the test is 30 points.

  1. 4 points Without using your calculator find the remainder when 20072007 is divided by 13.

2007 ≡ 707 ≡ 57 ≡ 5 (mod 13) 2007 ≡ 807 ≡ 87 ≡ 3 (mod 12) 20072007 ≡ 512 ·s+3^ ≡

( 512

)s · 53 ≡ 1 s^ · 25 · 5 ≡ − 5 ≡ 8 (mod 13)

So the remainder is 8

  1. 4 points Find the two smallest positive solutions to

2 x ≡ 8 (mod 4) 3 x ≡ 9 (mod 5) 4 x ≡ 10 (mod 6) Our system is equivalent to x ≡ 0 (mod 2) x ≡ 3 (mod 5) x ≡ 1 (mod 3) So x = 2a 2 a ≡ 3 ≡ 8 (mod 5) a ≡ 4 (mod 5) a = 4 + 5b x = 2(4 + 5b) = 8 + 10b 8 + 10b ≡ 1 (mod 3) b ≡ − 7 ≡ 2 (mod 3) b = 2 + 3c x = 8 + 10(2 + 3c) = 28 + 30c The two smallest positive solutions are 28 and 58

  1. 3 points Theorem: If a|c and b|c and if (a, b) = 1, then ab|c

Proof:

  1. c = as, c = bt definition of division
  2. ax + by = 1 for some x, y Since (a, b) = 1
  3. cax + cby = c Multiply both sides by c
  4. btax + asby = c substitute from 1.
  5. ab(tx + sy) = c ROF, factor out ab
  6. ab|c definition of division

In the boxes fill in a short reason for that step. An acceptable reason is ROA (rules of algebra).

  1. 4 points Characterize those positive integers n for which φ(n) 2

is odd. Clearly indicate your answer.

φ(4) 2

= 1. If n = pα^ for some odd prime p, then φ(n) = n

( 1 − 1 p

) = n

( p − 1 p

) .

So if

φ(n) 2 is odd,^4 does not divide^ p^ −^1 or^ p^ ≡^3 (mod 4) φ(2) = 1. So if (n, 2) = 1, then φ(2n) = φ(n)

n = 4 or if n = pα^ or n = 2pα^ with p prime and p ≡ 3 (mod 4), then

φ(n) 2 is odd.

MATH 302 Test 2 Solutions November 2, 2007 Page 2 of 2

  1. 4 points Let p be a prime number. Prove that if m = 2p^ − 1 is a composite number, then m is a pseudoprime number.

Since p is prime, 2 p^ ≡ 2 (mod p); i.e. p| 2 p^ − 2. We know that if m|n, then 2 m^ − 1 | 2 n^ − 1 , so 2 p^ − 1 | 22 p−^2 − 1. Consequently 2 p^ − 1 | 2 · (2^2 p−^2 − 1) = 2^2 p−^1 − 2 or 2 m^ ≡ 2 (mod m). Since m is composite, it is a pseudoprime.

  1. 4 points Prove: If (a, b) = 1, and (a, c) = 1, then (a, bc) = 1

ax + by = 1 and at + cs = 1 so 1 = (ax + by)(at + cs) = a^2 xt + axcs + byat + bcys = = a(axt + cxs + byt) + bc(ts) So (a, bc) = 1

  1. 4 points Use Wilson’s Theorem to find the remainder when 2(26!) is divided by 29

We know that 27! ≡ 27 · 26! ≡ − 2 · 26! ≡ 1 (mod 29) So 2(26!) ≡ − 1 ≡ 28 (mod 29) or the remainder is 28

  1. 3 points Use Fermat’s Theorem to show that for any prime p and integers a and b,

(a + b)p^ ≡ ap^ + bp^ (mod p)

Using Fermat’s Little Theorem three times we have a ≡ ap^ (mod p), b ≡ bp^ (mod p) (a + b)p^ ≡ a + b ≡ ap^ + bp^ (mod p)