apologize for the long post, please make corrections as it is likely that I've made mistakes
half of this is hindsight wisdom for me
q 26)
the number is 6037 and that will make the answer 16.
start with 10 keep adding 21 until the second condition is satisfied, then add 21*23 until the third condition is satisfied too.
q 27)
quite straight forward.
I think it's D, 5*sqrt(2)/3
q28)
cube both sides, rearranage, and that should give you 2 real roots only , 9 and -9
q30)
I think it's A, 364:
let the set be S.
it is not difficult to see that the only elements will be in the form 2^a * 3^b * 5^c (else you can keep taking out factors of 2,3, and 5 to get something not divisible by the numbers)
and X(k) be a subset of S for k = 0,1,2,3,4......
such that it contains all elements of S in the form (2^a * 3^b * 5^c) for a+b+c = k. a,b,c E Z+
for example
X(0) = {1}
X(1) = {2,3,5}
X(2) = {4,9,6,25,6,10,15}
now 2 points needs to be established:
1.) if X(k) is not empty then X(k) contains all numbers that can be written in the form in the form (2^a * 3^b * 5^c) where a+b+c=k.
X(k) is not empty .. so there is atleast 1 element
2^x * 2^y * 2^z.
by the conditions of the question, that will imply
2^x-1 * 2^y+1 * 2^z
2^x-1 * 2^y * 2^z+1
2^x * 2^y-1 * 2^z+1
...
are all in X(k) and by repeating the same process a number of times, we can get any number that could possibly in X(k).
2.) the size of X(k) is [(k+2) C 2 ]
consider a,b,c in 2^a * 2^b * 2^c.
starting from 'a'
and we can either
1.) move (to b)
2.) add one to current variable
we have k adds and 2 moves (a->b and b->c)
so together there are k+2 actions, and 2 of them are moves
so size of X(k) is [(k+2) C 2]
so the size of S must be X(0) + X(1) + .....
= 2C2 + 3C2 + .... nC2
and turns out to be 364 with n=13
