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·53≡1s·25 ·5≡ −5≡8 (mod 13)
So the remainder is 8
2. 4 points Find the two smallest positive solutions to
2x≡8 (mod 4) 3x≡9 (mod 5) 4x≡10 (mod 6)
Our system is equivalent to
x≡0 (mod 2) x≡3 (mod 5) x≡1 (mod 3)
So x= 2a2a≡3≡8 (mod 5) a≡4 (mod 5) a= 4 + 5b
x= 2(4 + 5b) = 8 + 10b8 + 10b≡1 (mod 3) b≡ −7≡2 (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 1−1
p!=n p−1
p!.
So if φ(n)
2is odd, 4does not divide p−1or p≡3 (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 p≡3 (mod 4), then φ(n)
2is odd.
