# Adv Alegbra HW Solutions 35 - ber of intervening days is...

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35 1.92 Find the smallest positive integer which leaves remainder 4, 3, 1 after dividing by 5, 7, 9, respectively. Solution. That the desired integer x satis f es three congruences: x 4mod5 ; x 3mod7 ; x 1mod9 . By the Chinese remainder theorem, the f rst two congruences give x 24 mod 35 . Now use the Chinese remainder theorem for the system x 24 mod 35 x 1mod9 (which is possible because ( 35 , 9 ) = 1. We obtain x 199 mod 315. Thus, 199 is the smallest such solution. 1.93 How many days are there between Akbal 13 and Muluc 8 in the Mayan tzolkin calendar? Solution. Akbal is month 3 and Muluc is month 9. If x is the intervening number of days, then x 13 8 mod 13 x 3 9 mod 20 . The Chinese remainder theorem gives x 174 mod 260 (and so the num-
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Unformatted text preview: ber of intervening days is either 174 or 86). 1.94 (i) Show that ( a + b ) n ≡ a n + b n mod 2 for all a and b and for all n ≥ 1. Solution. If a is even, then a + b ≡ b mod 2 and ( a + b ) n ≡ b n ≡ a n + b n mod 2 for all n ; a similar argument holds if b is even. If both a and b are odd, then a ≡ 1 ≡ b mod 2; hence, a + b ≡ 1 + 1 ≡ 0 mod 2 and ( a + b ) n ≡ 0 mod 2, while a n + b n ≡ 1 + 1 = 0 mod 2. (ii) Show that ( a + b ) 2 ±≡ a 2 + b 2 mod 3. Solution. If a = 1 = b , then ( a + b ) 2 ≡ 4 ≡ 1 mod 3, while a 2 + b 2 ≡ 2 mod 3. 1.95 Solve the linear system x ≡ 12 mod 25 x ≡ 2 mod 30 . Solution. Absent....
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## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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