View Full Version : More powerprep Question

Mystic

11-08-2004, 06:48 AM

Umm.......nobody answering my questions :(

OK here is one more , this is also from powerprep.

Hope I get this tonight

If n is an integer and 99<sqr(n)<200 then n could have at most how many values ?

() Two

() Four

()Six

()Eight

()Ten

Regards

M

calm_J

11-08-2004, 07:41 AM

hi

n =2

this is found from

99<n^2<200 then we can say 100<= n^2 < 200

then

10 <= n < 10 sqrt (2) this gives one value and

- 10 >= n > - 10 sqrt(2) this gives another value

bbm833

11-08-2004, 08:42 AM

You can calculate the square of numbers from 10 until you reach a number that is bigger than 200 :

10^2 = 100

11^2 = 121

12^2 = 144

13^2 = 169

14^2 = 196

15 ^2 = 225

as (-n)^2 = n^2

negative numbers should be considered.

Result set is -14, -13, -12, -11, -10, 10, 11, 12 , 13, 14

-->Answer is 10.

calm_J

11-08-2004, 09:01 AM

ohhh ooopppsss i missed to multiply 10* sqrt(2)

this means 14.1

whch gives 10 No.

Mystic

11-08-2004, 06:18 PM

Thanks!!!!!!!!

Got 10 also.

As before:

10^2

.

.

.

19^2

is 10 integers, so n =10