The product of all the factors of a number N is equal to the number raised to the power of four. If N is less than 100 then find the possible values of N .( A. 5 B. 6 C. 9 D .10)

Question

The product of all the factors of a number N is equal to the number raised to the power of four. If N is less than 100 then find the possible values of N .( A. 5  B. 6  C.  9  D .10)

ganeshie8 · Answer

As a start, notice that if $d$ is a factor of $N$, 
then $\dfrac{N}{d}$ is also a factor of $N$.

ganeshie8 · Answer

As $d$ runs over all the factors of $N$, 
the expression $\dfrac{N}{d}$ also runs over all the factor of $N$

ganeshie8 · Answer

Consider the product of all the factors $d$ and $N/d $ :

$$ \left(\prod d \right)\left(\prod \dfrac{N}{d} \right) =  \left(\prod d*\dfrac{N}{d} \right)= \left(\prod N \right) = N^{\tau(N)} $$

ganeshie8 · Answer

where $\tau(N)$ is the number of factors of $N$

ganeshie8 · Answer

basically you need to solve the equation : 

$$N^4 = N^{\tau(N)/2}$$

ganeshie8 · Answer

that is same as solving
$$\tau(N) = 8$$

ganeshie8 · Answer

how many positive integers less than 100 have exactly 8 factors ?

debpriya · Answer

Yes got it ...so after that how will we find the numbers less than 100 having exactly 8 factors ?

ganeshie8 · Answer

yea that looks tricky..

ganeshie8 · Answer

Consider the prime factorization of $N$

ganeshie8 · Answer

Would you agree that $N$ cannot have more than $3$ different prime factors

debpriya · Answer

Yes yes

ganeshie8 · Answer

so N can be of below forms : 

$2^a3^b5^c$

such that $a+b+c=8$

debpriya · Answer

But why only 2,3 5, ? Why cant we take 5,7,11?

ganeshie8 · Answer

Oh right, we should consider all small primes

debpriya · Answer

Oh the product will be more than 100!

debpriya · Answer

Got it ! Thank you so much for your help !

ganeshie8 · Answer

may i see how you work the final answer

debpriya · Answer

8 can be written as 1+7 , 2+6, 3+5, 4+4 which can also be  1+6+1,1+2+5,1+3+4,1+1+5,1+4+2,3+4+1,3+2+3,2+1+5,2+2+4 ?

debpriya · Answer

Is there a shorter method ?

ganeshie8 · Answer

I made a mistake

ganeshie8 · Answer

$N$ can be of form
$2^a3^b5^c7^d11^e$

such that $(a+1)(b+1)(c+1)(d+1)(e+1)=8$

right ?

ganeshie8 · Answer

because we want the number of factors to be equal to $8$
not the exponents of primes

ganeshie8 · Answer

here are the possible N values
 [24,30,40,42,54,56,66,70,78,88]

debpriya · Answer

Apologies for late reply since internet connection is slow

debpriya · Answer

Why have we considered 5,7,11 because won't the product be greater than 100 and doesn't the question mention that N is lesser than 100 ?