Question

the product of all the factors of a number N is equal to the tenth power of the number. which of the following cannot be the factor of N? (a. 90 b.200 c.210 d. 126)?

Answer 1

Anything to the tenth power ends in zero. So cross D out.

Answer 2

Oops. I read the question wrong. The answer is D. My bad.

Answer 3

5^10= 9765625
does not end in zero

Answer 4

3^10 does not end in zero either

Answer 5

5 or 0

Answer 6

3^10= 59049
ends in 9

Answer 7

the number of factors will be 20. that is the first step I know but I don't know how to do it further.

Answer 8

I give up. You can take it from here.

Answer 9

The prime factorization of 90 is 2*3*3*5

So the factors of 90 are
2, 3,3,5
2*3 , 2*3, 2*5 , 3*3, 3* 5, 3* 5
2*3*3, 2*3*5, 3*3*5
2*3*3*5

delete repeats
2,3,5
2*3, 2*5 , 3*3,3* 5
2*3*3, 2*3*5, 3*3*5
2*3*3*5

The product of the factors of 90 is
2*3*5*(2*3)*(2*5)*(3*3)*(3* 5)*(2*3*3)*(2*3*5)*(3*3*5)*(2*3*3*5)= 531441000000
90^10= 34867844010000000000

Answer 10

we can rule out 90

Answer 11

i saw one more solution where they found the number of factors of N. now there is a formula where the product of the factors is equal to N^m/2 where m is the number of factors .....now since in this question N^10 so the number of factors is 20. then they found the prime factorization of 20 which is 2*2*5. so they said that N can be written as a product of three prime factors. now since all the options except 210 have 3 prime factors we have to rule out 210 (it has 4 prime factors)...i did not get the logic in this ..but the answer is 210

Answer 12

I'm doing it out by brute force, and not getting that answer

Answer 13

factors of 210

2,3,5,7
(2*3), (2*5), (2*7),(3*5), (3*7),(5*7)
(2*3*5), (2*3*7), (2*5*7), (3*5*7)
(2*3*5*7)

the product of all these is 3782285936100000000, which is not equal to 210^10

did i leave out a factor of 210 ?

Answer 14

Heyy

Answer 15

nope you did not leave a factor of 210
answer is 210...i m just wondering if there is a shorter way to do this

Answer 16

I see, I misread the question. 210 is not N , it is a factor of N.

Answer 17

I am curious to find this number N .

Answer 18

i think N can have a lot of values

Answer 19

the simplest case , a prime number, is impossible p = p^10

a number with two factors, a,b,

a*b*(a*b) = (a*b)^10

Answer 20

Let \(d\) be a factor of \(N\), then we have
\[n = d*\dfrac{n}{d}\tag{1}\]

As \(d\) ranges over all the \(\tau(n)\) factors of \(N\), we get \(\tau(n)\) such equations.
Multiplying them all together gives
\[n^{\tau(n)} = \prod\limits_{d\mid n}d* \prod\limits_{\frac{n}{d}\mid n}\frac{n}{d}\tag{2}\]

Answer 21

But as \(d\) runs over the divisors of \(n\), so does \(\dfrac{n}{d}\); hence, \( \prod\limits_{d\mid n}d = \prod\limits_{\frac{n}{d}\mid n}\frac{n}{d}\)

Answer 22

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

which is same as
\[n^{\tau(n)/2} = \prod\limits_{d\mid n}d \]

Answer 23

We are given that the product of divisors equals \(n^{10}\) :

\[n^{\tau(n)/2} = \prod\limits_{d\mid n}d=n^{10}\]

Answer 24

so we endup solving \[\tau(n) = 2*10 = 20\]

Answer 25

Looks i have used \(n\) and \(N\) interchangebly...

Answer 26

In order to have \(20\) factors, \(n\) can only be of below forms :

\(p^{1}q^1r^{4}\)

\(p^3q^4\)

\(p^1q^{11}\)

\(p^{19}\)

Answer 27

Clearly \(n\) cannot have more than 3 different primes in its prime factorization

Answer 28

so none of the factors of \(n\) are allowed to have more than \(3\) different prime factors

Answer 29

In order to have \(20\) factors, \(n\) can only be of below forms :

\(p^{1}q^1r^{4}\)

\(p^3q^4\)

\(p^1q^{\color{red}{9}}\)

\(p^{19}\)

Answer 30

are you including 1 as a factor

Answer 31

yes all positive factors

Answer 32

Btw, above work makes sense only if you know how to find the number of divisors of a given integer

Answer 33

Let \[n = {p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots {p_r}^{e_r}\]

Do you know how to find the value of number of positive divisors \(\tau\) of \(n\) ?

Answer 34

@debpriya

Answer 35

Supplemental question.
Find the smallest number N such that the product of all the factors of N is equal to the tenth power of the number.

Answer 36

A couple of comments.

\( \large \prod\limits_{\frac{n}{d}\mid n}\frac{n}{d}\) is the same as \(\prod\limits_{d\mid n}d \), by listing the divisors backwards.

And you proved an interesting result. $$n^{\tau(n)/2} = \prod\limits_{d\mid n}d$$ http://primes.utm.edu/glossary/xpage/tau.html

Answer 37

@ganeshie8 i understood it. Thank you so much :)

Answer 38

Np :) see if you can find the smallest N whose product of divisors equals the 10th power of N