Find all numbers for which the product of all its factors (other than itself) is equal to the number itself.
49suppose 4:
in it shall we consider two 2s as factor or only one 2???
1aise to it ll be ne no. accept perfect squares nd no.s like 5 ,7 ,13............if m gettin the ques r8ly
62subhomoy, we count the factors only once....
chintan: that is one answer.. but how about the rest?
Sanchit: Din get u....
49oops.! :P
I initially read product of the other factors is half the number!
The answer is product of any 2 prime numbers exception being cubes of prime numbers! :)
21Let the canonical factorization of n be p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}}...p_{n}^{\alpha _{n}}
number of it's divisors is \tau (n) = (\alpha _{1}+1)(\alpha _{2}+1)..(\alpha _{n}+1)
We have \prod_{d|n}^{}{}d = n^{\frac{\tau (n)}{2}}=n^{2} \Leftrightarrow \tau (n) = 4. So n must be product of two primes or cube of a prime.
30Shubodip : Can you explain the step where you write \prod_{d|n}{d}=n^{\frac{\tau (n)}{2}}?
1@ ashish
n=p_1^{\alpha_1}.p_2^{\alpha_2}.p_3^{\alpha_3}\cdots \\ \text{Any divsior of n can be written in the form of } \\ d=p_1^{\alpha_i}.p_2^{\alpha_j}.p_3^{\alpha_k}\cdots \\ 0\leq \alpha_i\leq \alpha_1 \\ 0\leq \alpha_j\leq \alpha_2 \\ 0\leq \alpha_k\leq \alpha_3 \\ . \\ . \\ \text{So }\\ \prod{d}=p_1^{\left( 1+2+\cdots+\alpha_1\right)(\alpha_2 +1).(\alpha_3 +1)...}.p_2^{(\alpha_1 +1 )\left( 1+2+\cdots+\alpha_2\right).(\alpha_3 +1)...}.\cdots \\ \prod{d}=\left( p_1^{\alpha_1}.p_2^{\alpha_2}\cdots\right)^{\left(\frac{\left(1+\alpha_1 \right).\left(1+\alpha_2 \right).\left(1+\alpha_3 \right)...}{2} \right)}\\ \prod{d}=n^{\frac{\tau(n)}{2}} \\ \texttt{but this product also involves n as it is one of the permutation } \\ \texttt{So as per the given question,we have the product excluding itself is equal to n }\\ \text{Hence} \\ \frac{n^{\frac{\tau(n)}{2}}}{n}=n \\ \tau(n)=4
21yes. that is one proof. shorter proof
suppose n is not a perfect square. Let d be one of its divisor. So is \frac{n}{d} . their product is n. So we get \prod_{d|n}^{}{}d = n^{}\frac{\tau (n)}{2}. If n is a perfect square \sqrt{n} is a factor. So \prod_{d|n}^{}{}d = n^{}\frac{\tau (n)}{2}= n^{\frac{\tau (n)-1}{2}}\sqrt{n} = n^{}\frac{\tau (n)}{2}
