Question

find remainder when, 2¹⁹⁹⁰÷ 1990 using congruence

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Answer 1

Answer:

1990=2⋅5⋅199 .

21990≡0(mod2).

21990=(24)497⋅22≡4(mod5).

21990=(2198)10⋅210≡1024≡29(mod199).

Hence

21990=199k+29≡4(mod5)

⇒

−k−1≡4(mod5)

⇒

k≡0(mod5)

k=5p

⇒

21990=995p+29≡0(mod2)

⇒

p+1≡0(mod2)

⇒

p≡1(mod2)

⇒

p=2q+1

⇒

21990=995(2q+1)+29=1990q+102