Question

It is given that \(n\) is not a perfect cube.
If \(\prod\limits_{d\mid n} d = n^2\), then find the value of \(\prod\limits_{d\mid n^2}d\)

Options :
a) n^3
b) n^4
c) n^6
d) n^8
e) n^9

Answer 1

is this the answer->n^4 ?

Answer 2

nope

Answer 3

may i see your work ? it could be some arithmetic error...

Answer 4

\(\prod\limits_{d\mid n} d\)

means : "product of all the positive divisors of n"

Answer 5

example : n = 10

\(\prod\limits_{d\mid 10} d = 1*2*5*10 =100 \)

Answer 6

I got it

Answer 7

All numbers that divide n must also divide n^2

and i realized what mistake i made
i thought the greatest divisor of n^2 other than it will be n and then for \(\prod\limits_{d\mid n^2}d\) we just have an extra n^2 so we multiply it with\(\prod\limits_{d\mid n} d \) and get n^4

but my new approach->
is a divides n then a^2 will divide n^2

so divisors of n^2 will be all the divisors of n+ the square of all these divisors
so \[\prod\limits_{d\mid n^2} d =\prod\limits_{d\mid n} d \times \left( \prod\limits_{d\mid n} d \right)^2\]\[\prod\limits_{d\mid n^2} d=n^2 \times n^4\]\[\prod\limits_{d\mid n^2} d=n^6\]

Answer 8

that is a very nice try, but the answer is still wrong

Answer 9

consider an example

Answer 10

\(n = 2*3\)

the divisors of \(n\) are \(\{1,2,3,6\}\)
the divisors of \(n^2\) are \(\{1,2,3,6,4,12,9,18,36\}\)

notice that there are few extra divisors in \(n^2\).
In your proof, you're not counting these divisors of \(n^2\) : \(\{12, 18\}\)

Answer 11

ok ok i got where i was wrong as again

if we take some number say n
the prime factorization will be like-\[n=2^a 3^b5^c.....\]total factors will be given by all combination of multiplication of the factors
so total multiples will be->

\(No.~Of~Multiples~Of~n=(a+1)(b+1)(c+1)....\)

if we square n then we get prime factorization like this-\[n^2=2^{2a}3^{2b}5^{2c}.\]
\(total~multiples~of~n^2=(2a+1)(2b+1)(2c+1).....\) which is proves that there will be more factors

Answer 12

when you say "multiiple", you surely mean "divisors"

right ?

Answer 13

sorry
yes it is divisors

Answer 14

so 3 of the options are eliminated :)

Answer 15

i think we should use the information that n is not perfect cube

Answer 16

yes

Answer 17

So we got the answer n^9

Answer 18

that is the correct answer

Answer 19

BTW answer is not so important

Answer 20

The way is more

Answer 21

If we take n=8 which is a perfect cube
What will be the problems....

Answer 22

1*2*4*8 are 8 divisor
And their product is also 8 square .

Answer 23

Divisor of 64 are 1*2*4*8*16*32*64
And their product is 8^7

Answer 24

That why we are not given with n=perfect cube
Because some divisor remains same that's why we get less number of divisor
than 36

Answer 25

i like this method of assumption, we get the answer real quick :D
we make proper cases

consider\[n=p_1 \times p_2\]where \(p_1~and~p_2\) are primes

its clear that n is not a perfect cube
the divisors of n will be->\(1,~p_1,~p_2,~n\)
so \(\prod\limits_{d\mid n} d = n^2\) holds good

now when we see n^2\[n^2=p_1 \times p_2 \times p_1 \times p_2\]we can easily see all diff divisors of n^2
all diff divisors-\(1,~p_1,~p_2,~n,~p_1^2,p_2^2,~p_1^2p_2,~p_1p_2^2,~n^2\)

multiplying them we get->\((p_1p_2)(p_1p_2)^{2}(p_1p_2)^3(n)(n^2)\)=\(n^9\)

Answer 26

brilliant!

Answer 27

but this is not general solution because i made it up

Answer 28

you can make it general

Answer 29

you just need to prove below lemma first

\[\large n^{\tau(n)} = \left(\prod\limits_{d\mid n}d \right)^2\]

Answer 30

that restricts the number of prime factors to 2 as you're doing in your proof

Answer 31

Am also thinking about this method but I can't find the number of prime factor.
How can you say that they are only 2
It luckily coincide with 6 which is having only two primfactors.

Answer 32

http://prntscr.com/97lwd2

Answer 33

that will do :)

Answer 34

im sure this identity isn't hard for you to prove
\[\large n^{\tau(n)} = \left(\prod\limits_{d\mid n}d \right)^2\]

Answer 35

that conforms that \(n\) must have exactly two prime factors

Answer 36

\(\tau(n)\) = number of positive divisors of \(n\)

Answer 37

oh thats neat

Answer 38

\[\large n^{\tau(n)} = \left(\prod\limits_{d\mid n}d \right)^2\]

take srt both sides and get

\[\large n^{\tau(n)/2} =\prod\limits_{d\mid n}d \tag{1}\]

we're given \(\prod\limits_{d\mid n}d = n^2\tag{2}\)

comparing (1) and (2) gives
\[\tau(n)=4\]

Answer 39

how to generalize

Answer 40

\(\tau(n)=4\) means \(n\) is of form \(p^3\) or \(p_1p_2\)

Answer 41

because
the number of divisors of \(n=p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}\) is given by
\((e_1+1)(e_2+1)\cdots(e_r+1)\)

Answer 42

we can get \(\tau(n)=4\) only in two ways :

(3+1)

(1+1)(1+1)

Answer 43

ok yup

Answer 44

and we're given that \(n\) is not a perfect cube, so we can conclude \(n\) is a product of two distinct primes

Answer 45

im okay with that part, i mean how to generalize for all t(n)

Answer 46

rest is your/elusiveavenger's proof

Answer 47

i can see an induction way to build up a multiple of cases, or try and work out
t(n)=k

d|k of form (e1+1)(e2+1)....(en+1)

Answer 48

this would give us how many perfect squares and distinct primes we are dealing with

Answer 49

but its just expressions not closed form, the general case for all t(n) looks like work

Answer 50

we have showed that the given statement works only when \(n\) is a product of two distinct primes, right ?

Answer 51

ya

Answer 52

can u explain a bit more... what generalization ... i feel im not seeing something.. .

Answer 53

like this identity we saw to be true for t(n)=4 for once case, and that was because we saw how to move the d/n^2 case

Answer 54

but we still have to prove it really works for alll possible t(n) cases dont we

Answer 55

Oh you're referring to this identity ?
\[\large n^{\tau(n)} = \left(\prod\limits_{d\mid n}d \right)^2\]

Answer 56

i don't have a proof yet... need to think a bit... but im sure it wont be too hard..

Answer 57

il open a new question for this proof
it looks a bit interesting

Answer 58

okk

Answer 59

\[\frac{\tau(n)}{2} = 2\]
n has 4 divisors, in other words \(n=p*q\) or \(n=p^3\) BUT n is not a perfect cube!