Discrete Math Homework?

Question

1. Let n=2^14 * 3^9 * 5^8 * 7^10 * 11^3 * 13^5 * 37^10.
a) how many divisors does n have?
b) how many divisors in a are divisible by m=2^3*3^4*5^7*11^2*37^2
(where n^m is n to the mth power and * = times)

2. Let m=8408400 and n=9316125
a) write m and n as a product of primes.
b. Without using the Euclidean Algorithm, find the greatest common divisor of m and n

3. Prove n is odd if and only if n^2+1 is even

4. Disprove: If n, m are perfect squares then n+m must be a perfect square

5. Use the Fundamental Theorem of Arithmetic to find the smallest interget n so that n * 1260 is a perfect cube




NOTE: i KNOW it's kind of scrambled and hard to read, but I didn't know what else to do. this is seriously killing me and the professor is impossible to understand in class.

Answer 1

1 a. 15*10*9*11*4*6*11. Why? Because there are 15 choices for the exponent of 2 (namely 0, 1, 2, 3, ...., 15), 10 choices for the exponent of 3, 9 choices for the exponent of 5, and so on.

1 b. Now there are 12 choices for the exponent of 2 (namely 3, 4, 5, ... 14), 6 for the exponent of 3, and so on.

2 a. 8408400 is clearly divisible by 100=2^2*5^2. It has a further factor of 84, which you can factor as 2^2*3*7. What's left over is 1001, which I happen to know is divisible by 7 (because 980 and 21 both obviously are, and therefore also their sum 1001). I'll let you finish up the factorization.

n is obviously the sum of a number divisible by 1000 and hence 125, added to 125, and thus divisible by 125=5^3 itself. Also, the sum of its digits is divisible by 9, so it is too. Again, I'll let you finish up from there.

3. n is odd if and only if n^2 is. n^2 is odd if and only if n^2 + 1 is even.

4. 1 and 4 are perfect squares. Their sum, 5, is not a perfect square.

5. Factor 1260. You get 2*5*2*63 = 2^2*3^2*5*7. The problem is pretty equivalent to finding the smallest multiple of 1260 that's a perfect cube, and the latter number would be 2^3*3^3*5*3*7^3.

Answer 2

a million. b/a * b/c = b^2/ ac so that's pretend ex. b=2 ,a =3 , c=4 then 2/3* 2/4 = 4/12 = a million/3 no longer 2/12 = a million/6 2. b/a*c/a = bc /a^2 so that's pretend ex. b=2 ,a =3 , c=4 2/3*4/3 = 8 /9 no longer 8 /3 clean 3. b/a*d/c = bd/ac that's genuine ex. b=2 ,a =3 , c=4 , d = 5 2/3*5/4 = 10 / 12 that's an analogous as 2*5 / 3*4 = 10/12 solutions