Homework #5 7 th week Math 240 Thursday October 24, 2013

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1 . Let a, b > be integers and g : = gcd(a, b) its greatest common divisor. Show that if a = g q a and b = g q b then q a and q b are relatively rime. Since gcd(κ a, κ b) = κ gcd(a, b) in articular, for κ = g we have g = gcd(a, b) = gcd(g q a, g q b ) = g gcd(q a, q b ) gcd(q a, q b ) = that is, q a and q b are relatively rime. 2. Show that for any air of non negative integers a and b a b = gcd(a, b) lcm(a, b). Suose first that a and b are relatively rime and let m be any multile of both a and b. Then, for some integers q a and q b, m = a q a = b q b and so, a b q b. Since a and b are relatively rime it follows that a q b, i.e., q b = κ a for some integer κ which imlies that m = a b κ and hence a b m. This means that a b being a multile of a and b, is is a divisor of any its common multiles. Therefore, by the very definition of the least common multile, it follows that a b = lcm(a, b). Finally, if a and b were not relatively rime, writing a = g q a and b = g q b as in exercise, since q a and q b are relatively rime we have for we just have roved q a q b = lcm(q a, q b ) a b = (g q a )(g q b ) a b = g 2 lcm(q a, q b ) a b = g lcm(g q a, g q b ) a b = g lcm(a, b) a b = gcd(a, b) lcm(a, b) since as in exercise, g = gcd(a, b).

2 3. Find gcd(000, 625) (a) using the Euclidean Algorithm and (b) by factorization. (a) Successive divisions give the remainders 000 = = = = This means that the last non zero reminder is 25 and hence gcd(000, 625) = 25. (b) Since the rime factorizations of 000 and 625 are and 000 = = 5 4 we find that gcd(000, 625) = = 5 3 = (a) If is rime, show that the largest ower of dividing n! is or log n j= j = n n n n σ (n) where σ (n) denotes the sum of the base digits of n. (b) 000! has a lot of final zero digits. Use (a) to find how many are there. 2

3 (a) There are # { κ / κ, and κ n } = # { / κ κ n } = multiles of which are n. In the same way, for j =, 2,... there are # { κ / κ, and κ j n } { / = # κ κ n } j = j multiles of j which are n. Therefore, the largest ower of that divides n! is n n n Note that this sum ends u as soon as j > n, i.e., when j > log n. Alternatively, if n = a m m a m m a a 0 is the base exansion of n then, for any j =, 2,..., m, we have n j = a m m j a m m j a j a j a j a j a 0 j, but since 0 a i, a j a j a ( 0 () j j ) j ) ( j = () = j < we see that and hence j = a m m j a m m j a j a j m j= j = a m m a m m 2 a 2 a a m m 2 a m m 3 a 3 a 2. a m a m a m 3

4 = a a 2 ( ) a 3 ( 2 ) a m ( m ) = a () a 2 ( 2 ) a 3 ( 3 ) a m ( m ) ( a a 2 2 a 3 3 a m m) (a a 2 a 3 a m ) = = n σ (n). (b) If s (n) denotes either of the quantities aearing in art (a), the rime decomosition of n! is n! = n rime s(n). Since the number of zeros at the end of n! coincides with the largest ower of 0 = 2 5 dividing n! and s 5 (n) < s 2 (n) we see that the total of such zeros is s 5 (n). In articular, when n = s 5 (000) = = = and 000! ends with 249 zeros. 5. (a) Given two non negative relatively rime integers a an b, show that if x 0, y 0 is a articular solution of the Diohantine equation ax by = m then, any other solution is of the form { x = x0 bκ for some integer κ. y = y 0 aκ (b) Use (a) to describe the solution set for the general linear Diohantine equation ax by = m when a and b are arbitrary non negative integers. (a) If x 0, y 0 satisfies ax 0 by 0 = m and x, y is any other solution of this equation, i.e., ax by = m, by subtracting a(x x 0 ) = b(y y 0 ). This imlies that b a(x x 0 ) and hence b (x x 0 ) because a and b are relatively rime. This means that for some integer κ, x = x 0 bκ. Also, from the above relation it follows that b(y y 0 ) = abκ and so y = y 0 aκ. 4

5 (b) Let x 0, y 0 be a solution of the general equation ax by = m. We know that if g : = gcd(a, b) then, g m so if, as in exercise, we write a = gq a and b = gq b, any solution x, y to the equation will satisfy a a x q b y = m g Z. Since q a and q b are relatively rime (exercise ), from art (a) { x = x0 κq b for some integer κ. y = y 0 κq a 6. Solve } x mod 3 x 2 mod 5 From the first equation x = 3κ and from the second x = 2 5l form some integers κ and l. This means that for x to be a solution of the given system, κ and l must satisfy 3κ = 2 5l 3κ = 5l. Since 3 and 5 are relatively rime and κ 0 = 2, l 0 = is a articular solution to this last equation, we see that its solutions are describe (exercise 5) by { κ = 2 5υ l = 3υ where υ Z is an arbitrary integer. Thus, returning to the exression for x in terms if κ (or l) we find that the general solution to the given system of congruences is x = 7 5υ with υ Z an arbitrary integer. In other words (recall the Chinese reminder theorem), } x mod 3 x 7 mod 5. x 2 mod 5 5