HarveyS wrote:
For integers x, y, and z, x = y^2. What is the value of z?
(1) x = z!(z−1)!
(2) 12 < z < 22
There ought to be a subtype called "abstraction in question stem " subtype. whenever you see some serious math in the question stem, try some simplified versions of what is going on to try to uncover a pattern. Thats what the test taker is really wanting you to do
notice x = y^2 is there to tell us x is a perfect square.
(1)
Ok lets try some small values of z
if z = 3, Z!(z-1)! = 3!2! = 3x2x2 = 3*2^2
if z = 4, Z!(z-1)! = 4*3*2*3*2=4(2*3)^2 since 4 is perfect square, we have perfect square
if z=5 ................5*4*3*2*1*4*3*2*1 = 5(4*3*2*1)^2
if z = 6.............................................6(5*4*3*2*1)^2
if z = 7...........................................7(6*5*4*3*2)^2
if z = 8 ..........................................8(7*6*5*4*3*2)^2
if z = 9..........................................9(8*7*6*5*4*3*2)^2 a perfect square so we can notice two things fronm this little exercise
1) z!(z-1)! = z((z-1)!)^2)
2) n!(n-1)! will only be a perfect square only if z is a perfect square (1) is clearly NS
(2)
12<z<22 clearly NS
(1) and (2) Ok, between 12 and 22 there is only one perfect square. that number if 16. So, we are sufficient
OA is C