# Adv Alegbra HW Solutions 33 - x y and z such that x 2 y 2 z...

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33 1.82 (i) Prove that 10 q + r is divisible by 7 if and only if q 2 r is divisible by 7. Solution. If 10 q + r 0 mod 7, then 15 q + 5 r 0 mod 7, and so q 2 r 0 mod 7. Conversely, if q 2 r 0 mod 7, then 3 q 6 r 0 mod 7, hence 3 q + r 0 mod 7, and so 10 q + r 0 mod 7. (ii) Given an integer a with decimal digits d k d k 1 ... d 0 ,de f ne a 0 = d k d k 1 ··· d 1 2 d 0 . Show that a is divisible by 7 if and only if some one of a 0 , a 0 , a 0 ,...isdivisible by 7. Solution. If a = 10 b + d 0 , then a 0 = b 2 d 0 . By part (i), a 0 mod 7 if and only if a 0 0 mod 7. Now repeat. 1.83 (i) Show that 1000 ≡− 1 mod 7. Solution. Dividing 1000 by 7 leaves remainder 6 ≡− 1 mod 7. (ii) Show that if a = r 0 + 1000 r 1 + 1000 2 r 2 +··· , then a is divisible by 7 if and only if r 0 r 1 + r 2 −··· is divisible by 7. Solution. If a = r 0 + 1000 r 1 + 1000 2 r 2 +··· , then a r 0 + ( 1 ) r 1 + ( 1 ) 2 r 2 +···= r 0 r 1 + r 2 −··· mod 7 . Hence a is divisible by 7 if and only if r 0 r 1 + r 2 −··· is divisible by 7. 1.84 For a given positive integer m , f nd all integers r with 0 < r < m such that 2 r 0mod m Solution. The answer depends on the parity of m . 1.85 Prove that there are no integers
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Unformatted text preview: x , y , and z such that x 2 + y 2 + z 2 = 999. Solution. Now 999 â‰¡ 7 mod 8. But no sum of three numbers, with repeti-tions allowed, taken from { , 1 , 4 } , adds up to 7. This is surely true if a 4 is not used, while if a 4 is used, then the largest sum one can get which is under 8 is 6. 1.86 Prove that there is no perfect square a 2 whose last two digits are 35. Solution. If a is a positive integer, then a = d + 10 d 1 + 100 d 2 + Â·Â·Â· + 10 n d n , where 0 â‰¤ d i â‰¤ 9 for all i . Therefore, a â‰¡ d + 10 d 1 mod 100. In particular, the last two digits of a are 35 if and only if a â‰¡ 35 mod 100. Let b be a positive integer with b 2 â‰¡ 35 mod 100. Now the last digit of b must be 5 (otherwise the last digit of b 2 would not be 5), and so...
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