CPhill

The complement of x = (90- x)

The supplement of x = (180 - x)

(9x^5 + 25x^3 - 4) - (x^5 - 3x^3 - 4)  =

8x^5 + 28x^3   =

4x^3 (2x^2 + 7)

(6a^3 + 92a^2 - 7)  - (-7a^3 + a^2 -7)   =

13a^3 + 91a^2   =

13a^2 (a + 7)

Just distribute the "3" across the terms in parentheses....we get ......24x^2 - 6x  + 3

Here's a pic......

The area of this quadrilateral is just B x H  = (3 x 3)  = 9 sq units.......

Melody, I agree with your answer to (b).....I was assuming unique primes.....but clearly there could not be 4 unique primes that would produce a positive interger with exaxctly 20 divisors.....so .....your answer of 3 is correct.....

c)  I believe the answer to this is 240....to see why......

As per my explanation to  (b) above, the largest numer of primes that could factor such a  number is 4.....

Note that  2,3 ,5 and 7   are the smallest primes......then...using the reasoning from (b), above.... we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20  ...  where k, l ,m and n ≥ 1......so.....this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors...2, 3 and 5....again  we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have  2 *2 *5  = 20.....and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining posibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20.....clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648   or 2^4 * 3^3 = 432....and both are > 240......

b) .....I believe the answer to this is 4.....to see why......

Note that.....the number of possible divisors for a^k (where a is prime, and k ≥ 1)  is just (k + 1)

For instance, 3²  = 9, has  (2 + 1) = 3 possible divisors...... 1, 3 and 9

And the number of possible divisors for a^k * b^m    where a, b are prime, and k, m ≥ 1 .....  is just   (k + 1) (m +1)

So.....consider a positive integer factored as a^k * b^m *cⁿ * d^p  where a, b, c, d are prime and k, m, n and p  = 1

Then, the possible number of divisors is just (k + 1)(m + 1)(n + 1)(p + 1)  = (2)(2)(2)(2)  = 16

Note that, if another prime (also to the first power)  factored the number, we would have (2)(2)(2)(2)(2)  = 32 possible divisors.....and that's too  many.....!!!!

Edit...this answer is incorrect......Melody has the correct solution, below.....!!!!!