if the product of all the factors of a number N is equal to the square of the number and the sum of all the factors of the number other than the number itself is 21, then find the number of possible values for the number? a. 0 b.1 c.2 d. more than 2

Question

if the product of all the factors of a number N is equal to the square of the number and the sum of all the factors of the number other than the number itself is 21, then find the number of possible values for the number?                                                                 a. 0   b.1   c.2   d. more than 2

ganeshie8 · Answer

$n^{\tau(n)/2} = \prod\limits_{d\mid n}d = n^2$

That means $\tau(n) = 4$

debpriya · Answer

yes yes !

ganeshie8 · Answer

I'm using n = N

ganeshie8 · Answer

You familiar with the number theoretic functions $\tau$ and $\sigma$ ?

ganeshie8 · Answer

$\tau(n)$ is number of positive factors of $n$

$\sigma(n)$ is sum of positive factors of $n$

debpriya · Answer

yep so the way we can write the number will be p^1*^1 and p^3  right ? based on the previous question ?

debpriya · Answer

i meant p^1 * q^1  
where p and q are prime

ganeshie8 · Answer

Excellent!  only below forms are possible for $n$ to have exactly 4 positive factors :

$p^3$

$p^1q^1$

ganeshie8 · Answer

What are the factors when of $n$ when $n=p^3$ ?

debpriya · Answer

the factors will be only p right ?

ganeshie8 · Answer

factors of $n$ are the integers that divide $n$

debpriya · Answer

oh and p square ?

ganeshie8 · Answer

think a bit, there should be exactly $4$ positive factors of $p^3$ right ?

debpriya · Answer

so the factors are 1,p, p square and p cube ?

ganeshie8 · Answer

Yes!

ganeshie8 · Answer

what are the factors of $p^1q^1$ ?

debpriya · Answer

1,p,q and pq

ganeshie8 · Answer

so we need to solve
1 + p + p^2 + p^3 = 21

ganeshie8 · Answer

and independently

1 + p + q + pq = 21

ganeshie8 · Answer

Key things to keep in mind to arrive at the answer quick : 

1) 21 is an odd integer
2) sum of four odd integers is always even
3) 2 is the only even prime number

yttrium · Answer

What subject is this?

debpriya · Answer

there fore p+q+pq= even , thus p ,q and pq should also be even. am i right?

ganeshie8 · Answer

Also keep in mind p and q are two "different" primes

ganeshie8 · Answer

number theory @Yttrium

ganeshie8 · Answer

Do we have two different even primes ?

debpriya · Answer

nope

debpriya · Answer

so this wont work :(

ganeshie8 · Answer

so 1+p+q+pq = 21 has no solutions

ganeshie8 · Answer

How about the other equation
1 + p + p^2 + p^3 = 21

?

debpriya · Answer

here p+p^2+p^3 should be even ....so in this case p=2

ganeshie8 · Answer

Yes, plugin p=2 and see if it satisfies the above equation

debpriya · Answer

it doesn't satisfy the equation

ganeshie8 · Answer

can we conclude there exists no integer $n$ that satisfies the given conditions ?

debpriya · Answer

so answer is 0?

ganeshie8 · Answer

Yep!

debpriya · Answer

i don't know why but in the solution set they have marked it as 2. But this method did make sense

ganeshie8 · Answer

Oh, we made a mistake

ganeshie8 · Answer

the sum of all the factors of the number `other than the number itself` is 21

ganeshie8 · Answer

we should exclude the number itself from the sum, but we have added all the factors

ganeshie8 · Answer

we should be solving the equations : 

1 + p + p^2 = 21

and

1 + p + q = 21

ganeshie8 · Answer

try again

debpriya · Answer

in the first equation p has to be odd..coz 2 doesn't satisfy...

debpriya · Answer

but the first equation has no such odd number only the second equation can we satisfied by taking two odd prime number below 20 which are 3,17 and 7 ,13 therefore the answer is 2..am i right?

ganeshie8 · Answer

Sounds good to me!

debpriya · Answer

thank you once again ! You have been a great help :)

ganeshie8 · Answer

Np:)