x and x 2 is the divisor of an integer n If n has 4 divisors including 1 and itself

n has divisors 1,x,x+2,n. So this is possible if n= x(x+2)=x^2+2x,
Now, n+1=x^2+2x+1=(x+1)^2 ,is a perfect square.
7 years agoHelpfull: Yes(20) No(0)

Assume n = 8
4 divisors of 8 are 1, 2, 4, 8
Here x is 2 and x+2 is 4
Now n = 8
n+1= 9
9 is perfect Square
Ans : b
7 years agoHelpfull: Yes(13) No(3)
choose x=5
x+2=7
n will be 35
and 35+1 =36
perfect square
7 years agoHelpfull: Yes(5) No(0)
Let us take the 2 digit no i.e.., 15
4 divisors of 15 are 1,3,5,15
Here x is 3 and x+2 is 5
Now we got n =15;n+1=16
16 is perfect Square Ans: b
7 years agoHelpfull: Yes(4) No(0)
Then what about n=10
1,2,5,10
It's a prime number
7 years agoHelpfull: Yes(1) No(17)
checking for perfect squares ,orders are 1,4,9,16,25,36,....which would stand in (n+1) place, which leaves 0,3,8,15,24,35..... to be (n). 8 is suitable number with given input which has 1,8,2,4 as divisors satisfying x and x+2 condition as divisors
7 years agoHelpfull: Yes(1) No(0)
Assume r1=r2=0 then
n=q1*x
n=q2*(×+2)
x*q1=(x+2)*q2
For equality
q1=x+2 q2=x
so,
n=x*(x+2)
n+1=(x+1)^2
for all x n+1 is perfect square
7 years agoHelpfull: Yes(1) No(0)
Let n=8 and x=2 ; x+2=4
i.e. for n=8 we have 4 divisors 1, 2, 4 & 8.
so, n+1=8+1=9 is a perfect square.

Now, if we take n=12 and x=2 ; x+2=4
i.e. for n=12 we have 4 divisors 1, 2, 4 & 12.
so, n+1=12+1=13 is a prime number.

so, the answer is C
7 years agoHelpfull: Yes(1) No(3)
Assume n=8
Divisors are 1,2,4,8
then n+1=9 i.e. perfect square.

Again, assume n=10
Divisors are 1,2,5,10
then n+1=11 i.e. prime number
Thus, Answer: c
6 years agoHelpfull: Yes(0) No(0)

x and x+2 is the divisor of an integer n.If n has 4 divisors including 1 and itself.then n+1 will be
a) prime number
b)perfect square
c)all of the above