1.3 Integers Modulo n

a. True. 40−13=3·9
c. True. −29−6=(−5)7
e. True. 8−8=0·n for any n.
g. False. 84≡(64)2≡(−1)2≡1 (mod13).
a. 2k−4=7q, so q is even. Thus k=2+7x for some integer x; that is k≡2 (mod 7).
c 2k≡0 (mod 9), so 2k=9q. Thus 2 img q, so k=9x for some integer x; that is k≡0 (mod 9).
a. 10≡0 (mod k), so k img 10: k=2, 5, 10.
c. k2−3=qk, so k img 3. Thus k=1, 3 so, (as k≥2 by assumption) k=3.
a. ab (mod 0) means ab=q·0 for some q, that is a=b.
a. aa for all a because n img (aa). Hence if n img (ab), then n img (ba). Hence if ab=xn and bc=yn, img, then ac=(x+y)n.
7. If n=pm and ab(mod n), then ab=qn=qpm. Thus ab(modm).
a. In img, so , . Since 515=6·85+5 we get . Hence ...

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