shmewy

The smallest possible positive integer value of x would be 1, since 5 + 2*-2 =1. The answer to part one of your question is \(\boxed{1,-2}\)

im pretty sure the euclidean algorithm gives infinite solutions. after i did the process x=1+2k, y=−2−5k. so you just plug in values of k to get solutions. Im not sure what the question is asking but im pretty sure it wants \(\boxed{x=1+2k,\space y=−2−5k}\). 

well for an integer to have 8 positive divisors, it means that its prime factorization is either p*q^3 or p*q*r, where p,q, and r are primes. uhhh its literally imposible to determine all the positive integers with this prime factorization since there are infinite numbers of primes. :P

so i would start this problem by starting with a number like 1300 then getting rid of the squares and cubes less that 1300 and see how close i am. so 1300 is about 1296 or 36^2. 10^3 is the greatest cube less than 1300. so there would be 36+10 or 46 squares and cubes. but we overcounted the 6th powers. 1, 2 and 3 are in the range so we must substract 3. 43 in total. which means our initial guess is way too high. so ill try 1050. there are still 10 cubes but only 32 squares. so there are 39 values that work. so still too high. ill try 1030. still 10 cubes and 32 squares. now ill try \(\boxed{1039}\), and i think it works!!!

in my previous solution i got 5*4*3 /6 but now i think its 5*4*3/3 since you can rotate the equilateral triangle to only three sides. so im pretty sure its \(\boxed{20}\)?

to get the number of divisors an integer has we multiply the powers of the primes +1. for example 20 is 2^2*5^1 so the total number of divisors is (2+1)(1+1) or 6. so we need to reverse this process for 5. 5 is prime so it has to be 1*5 or p^4. there can only be \(\boxed 1\) prime that divides a terrific positive integer. so much for being terrific ;P

1+2+3...198+199+200 is 50*201 since you can pair up the first and last, 2nd and 2nd to last, etc. 

201 is 3*67 and 50 is 2*5^2. the greatest prime factor is \(\boxed{ 67}\) :P

ahhhhh... i was interpreting that mark had to be in the committee. Thanks :)