The positive integers are such that p

Question

The positive integers are such that p < q ≤ r < s < 100, ps = qr and √s - √p ≤ 1. What is the value of p?

(X) The last digit of s is either 1, 2 or 3
(Y) 50 < p and r < 90

amit2k9 · Answer

a s can be 1,2,3 means s can be 81 and p can be 64 giving s^(1/2) - p^(1/2) = 1
 similarly s and p can have values such that s^(1/2) - p^(1/2) <= 1 decimal values possible.

b tells nothing about p and s relatively.

a+b where s = 81, p = 64 q and r = 72 each fits perfectly.
hence C.

hussi9 · Answer

Solution:

Given that s-p > r -q as p < q <= r < s
=> (s-p)^2 > (r-q)^2
=> (s-p)^2 + 4sp > (r-q)^2 + 4rq [becuase sp = rq, we add 4sp in LHS and 4rq in RHS]
=> (s+p)^2 > (r+q)^2
=> s+p >= r+q+1 [as all numbers are integers] -> (1)

Suppose √s - √p = 1 (the other possibility is √s - √p < 1 that we will see later)
=> s + p - 2√sp = 1 => s+p = 1 + 2√qr [becuase sp = rq]
but (1) tells that √qr >= q + r => r = q [By AM-GM rule on positive numbers] and p+s = 2q + 1.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)
Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, x) = 1 and gcd(b, x) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1].

If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

=> In all s and p are perfect squares. Now take X -> only possible s is 81 => p = 64
Now take Y, p > 50 => p can be 64 or 81 but if p = 81 then s = 100 (not possible as s < 100) => p can only be 64. The information on r is required to cross-check if our data in hand is correct and it indeed is as √64.81 = 72.

IanStewart · Answer

Where is this question from? It's one of the most unreasonable practice questions I've seen. Completing the proof alone takes well longer than two minutes, and it also requires repeated application of the Arithmetic Mean/Geometric Mean inequality, which is beyond the scope of the GMAT. Unless you're applying to do a Masters degree in Number Theory, you should ignore this question and work on more realistic practice material.

Not only that, but the OA posted is incorrect. If you do the work you find that the only sequences p, q, r, s which satisfy all of the given conditions are in the form x^2, (x)(x+1), (x)(x+1), (x+1)^2 where x is an integer. Since that's the case, each statement alone forces our sequence to be 64, 72, 72, 81, so the answer ought to be D, not C.

The positive integers are such that p < q r < s <

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