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If n is a positive integer and is not the square of any [#permalink]
18 Feb 2006, 15:19

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

136. If n is a positive integer and is not the square of any integer, is it a prime number?
1). Among the factors of n, only n is greater than n^0.5
2). Among the factors of n, only 1 is less than n^0.5

136. If n is a positive integer and is not the square of any integer, is it a prime number? 1). Among the factors of n, only n is greater than n^0.5 2). Among the factors of n, only 1 is less than n^0.5

intresting.
suppose n = 9, n^0.5 = 3 and factors of n = 1, 3, 9.
suppose n = 7, n^0.5 = 2.3 and factors of n = 1, 7.

136. If n is a positive integer and is not the square of any integer, is it a prime number? 1). Among the factors of n, only n is greater than n^0.5 2). Among the factors of n, only 1 is less than n^0.5

intresting. suppose n = 9, n^0.5 = 3 and factors of n = 1, 3, 9. suppose n = 7, n^0.5 = 2.3 and factors of n = 1, 7.

136. If n is a positive integer and is not the square of any integer, is it a prime number? 1). Among the factors of n, only n is greater than n^0.5 2). Among the factors of n, only 1 is less than n^0.5

intresting. suppose n = 9, n^0.5 = 3 and factors of n = 1, 3, 9. suppose n = 7, n^0.5 = 2.3 and factors of n = 1, 7.

n = 9 and/or 7 works for both statements. so E.

but isnt N=9, a square of an integer (3)

good point. i overlooked. A seems correct now. but work later..

Seems like the root of integers (that aren't perfect squares) lies somewhere between the two middle factors of that integer.

sqrt 15 = 3,8

15: 5*3

sqrt 350 = 18,7

350: 2*5*5*7 or with factors 1,2,5,7,10,14,25,35,50,70,175,350

If the highest factor greater than the sqr is the number itself, it has to be a prime.
If the factor below the root is 1 then the integer has to be a prime.

A)Sufficient.
Let's consider that A is true and n is not prime.
We consider p the greatest factor of n different from n then we have:
n=p.q with 1<q<=p
=> n<=p^2
from A we have p<sqr(n)
=> n<sqr(n)^2
=> n<n impossible.
So n must be a prime number.

B)Sufficient.
We can use the same reasoning with q the the lowest factor of n.

If n is a positive integer and is not the square of any integer, is it a prime number?
1). Among the factors of n, only n is greater than n^0.5
2). Among the factors of n, only 1 is less than n^0.5

n = 2, 3, 5, 6, 7, 8, 11

let's pick 7 the factors are 1, 7, only 7 is greater than sqrt (7), which is true. 7 is prime.

other options are 6, 8, 14..which really don't satisfy the condition given in statement A.