1. If x^2-px-q=o, where p and q are positive integers, which of the following could not equal x^3?

A. 4x+3 B. 8x+5 C. 8x+7 D. 10x+3 E. 26x+5

 

2. X is a positive integer in which each digit is 1; that is, X is of the form 111...111. Given that every digit of the integer pX^2+qX+r (where p, q and r are fixed integer coefficients and p > 0) is also 1, irrespective of the number of digits X, which of the following is a possible value of q?

A. -2 B. -1 C. 0 D. 1 E. 2

 

Just give me some hints, then I can work them out. Thank you very much!

Edited November 7, 201015 yr by Simon Y

Howdy,

 

I know I'm extremely late in replying but I can't resist looking at the BMA questions

 

I don't have a huge amount of time to look, but the first one looks pretty doable. For instance, I noticed that if [math]x^2-px-q=0[/math] then [math]x^2 = px+q[/math] and so [math]x^3 = px^2 + qx = p(px+q) + qx = (p^2+q)x +pq[/math]. Now equate co-efficients to your possible solutions and see if you can figure out which one doesn't have solutions. Another hint: you don't have to solve five sets of simultaneous equations

Can't help but post the solution up here so others can see where I'm coming from. In the last post I got as far as any possible values of [math]x^3=(p^2+q)x+pq[/math]. In particular, notice that all of your possible solutions are of the form [math]ax+b[/math] and [math]b[/math] is prime. Equating co-efficients of the two, you must have [math]pq=b[/math] which therefore yields two solutions [math](p,q) = (b,1)[/math] or (1,B). If the candidate solution is possible, then evaluating [math]p^2+q[/math] should equal a for at least one of these combinations.

 

Quickly looking at the candidate solutions, you see that B requires [math]pq=5 \Rightarrow (p,q) = (1,5)[/math] or [math](5,1)[/math] and hence [math]p^2+q = 6[/math] or 26. Since neither of these equal 8, [math]x^3[/math] cannot be equal to [math]8x+3[/math].