Question

product of all factors of number N is equal to tenth power of the number . then N cant be divisible by 90 , 200 , 126 or 210?

Answer 1

We need to prove it ?

Answer 2

i give 4 numbers, you have to choose the correct option.
basically the question is:
product of all factors of number N is equal to tenth power of the number . then N cant be divisible by
a. 90
b. 200
c. 126
d. 210

Answer 3

Is the ans 210 ?

Answer 4

By my work, my best bet will be 210

Answer 5

well i dont have the solutions right now but i too got 210 by looking at its factor.
how did you get it?

Answer 6

Let N be non perfect square.
Let there be m factors of N, then product of all factors of N = N^(m/2)

We are given N^(m/2) = N^10

hence m=20
factoring 20 -> (1*20) = (2*10) = (4*5) = (2*2*5)
There can thus be max 3 prime factors of N
Only 210 has 4 prime factors, rest all have 3 . And hence the asn

Answer 7

why is N= N^(m/2)
and how do you know N is a non perfect square

Answer 8

im going to bed i will see you tomorrow

Answer 9

It is n^10 = n^(m/2)

Answer 10

where does m come from?

Answer 11

please explain more , thanks

Answer 12

And Its and assumption, you'll have to take another case where N is a perfect sq.

Answer 13

an*

Answer 14

Lets take an example, say 20

Answer 15

factors = 1,2,4,5,10,20

Answer 16

product = (1*20) * (2*10) * (4*5)

Answer 17

there are more, 8,

Answer 18

oh woops

Answer 19

60 is such a number... as long as factors don't include the number itself.

Answer 20

oh woops

Answer 21

so product = 20^3 = 20^(6/2)

Answer 22

Whats with "woops" today? o.O

Answer 23

Too impulsive... working on a solution :P

Answer 24

So N^(m/2) is clear ?

Answer 25

Trying a different approach... hang on.

Answer 26

this is true for all numbers?

Answer 27

Which are not squares.

Answer 28

Alternatively, all sq. no. have odd no. of factors, otherwise, all other nos. have even no. of factors.

Answer 29

So 2 of those factors will always multiply together to give N. No. of pairs = m/2
Hence product = N^(m/2)

Answer 30

thats not a rigorous argument, but ok, i guess this just popped up

Answer 31

What do you mean by not a rigorous argument ?

Answer 32

doing one example doesnt mean its true for all non perfect squares

Answer 33

Lets try to generalize.

Answer 34

Let factors of N be p1, p2,p3 .. pm (i.e. m factors)

Answer 35

Where p1<p2<p3..<pm

Answer 36

p1 * pm = N

Answer 37

None of which are 1 or N, right? You can be specific and let there be 2k factors, but

Answer 38

p2 * pm-1 = N

Answer 39

Nopes. 1 and N included.

Answer 40

If we exclude both, product will be N^(m-2)/2

Answer 41

well, fine, but you can be specific and make it so that they are
p1, p2,p3... pk,
And then the rest are
N/p1,N/p2,...N/pk

Like that

Answer 42

ok i think I see what you did

Answer 43

Glad you do.

Answer 44

(2^0 * 5^0) * (2^2 * 5^1
(2^1 *5^0) * ( 2^1 * 5^1) ...

Answer 45

?

Answer 46

how do you find N

Answer 47

We don;t have to find N, where did I find N

Answer 48

you wrote
There can thus be max 3 prime factors of N
Only 210 has 4 prime factors, rest all have 3 .

Answer 49

how did you reason that there can be max 3 prime factors

Answer 50

Yes.
m =20 , i.e. N has total 20 no. of factors.

Answer 51

20 can be written as
1*20
2*10
4*5

Answer 52

or 2*2*5

Answer 53

Before this, you should be knowing that , say
n = x^p . y^q . z^r ,
where x,y,z are prime nos, then total factors of n are (p+1)(q+1)(r+1)

Answer 54

im not sure what they are asking. you didnt find N

Answer 55

terenz, what are they asking?

Answer 56

Something tells me I should stay out of this, but it's only asking which of those four numbers cannot divide N.

Answer 57

We are not supposed to find N, we are supposed to find which of the following (90,126,210,200) is not a factor of N

Answer 58

ok

Answer 59

Are you following till now ?

Answer 60

Do you know about no. of factors thing ? Like I just said , (p+1)(q+1)(r+1) ?

Answer 61

now sure how you got that

Answer 62

hmmm..

Answer 63

nevermind, i see that

Answer 64

thats just using number of choices

Answer 65

you sure?

Answer 66

Yep.

Answer 67

yeah, im lost on the other stuff

Answer 68

so we deduced that N has 20 factors,
because N=N^(m/2) = N^(10) ---> m /2 = 10 , m = 20

Answer 69

yes.

Answer 70

20 must be of the form of (p+1)(q+1)(r+1)(s+1) ...

Answer 71

Where p,q,r,s.. are all natural numbers.

Answer 72

N = x^p . y^q . z^r *...
20 = (p+1)(q+1)(r+1) ... ok

Answer 73

We also know
20 = 1*20 = 2*10 = 4*5 = 2*2*5

Not any more combos possible right ?

Answer 74

right

Answer 75

Lets sat p<=q<=r<=s...
On comparison,
p can be 0 and then q = 19 (for 1*20)
p can be 1, and q = 9 ( for 2*10)
p can be 3 ,q=4 , (for 4*5)
p can be 1, q=1 and r =2 (for 2*2*5)

Answer 76

Heh... I give up @shubhamsrg
Your way was probably the only way that's not beyond my own capabilities...

There's no need to reinvent the wheel...

Answer 77

hmmm. I should take that as a compliment! :P

Answer 78

r = 2 ?

Answer 79

sorry 4*

Answer 80

ok, now how does that help to get the conclusion

Answer 81

I literally made guesses on possible combinations of the first three prime numbers (my rationale was that 3 is the greatest natural number where 3! < 10)

And figured out (experimentally, not logically) that (p+1)(q+1)(r+1) thing @shubhamsrg had already come up with earlier.

I guess the least possible value for N is 240.

Answer 82

let me think about this and come back tomorrow
you also have to look at the perfect square case?

Answer 83

Well, anyway, @shubhamsrg way of doing it had no flaw to it that I could spot, but apparently just needed more explaining :P

Answer 84

Now we know N = x^p . y^q . z^r
there is no possibility of a "s"

Answer 85

Hence N can only we represented in terms of 3 prime factors.
only 210 is the one which can not be represented by 3 prime numbers

Answer 86

@perl
Cannot be a perfect square if you accept that
N=(x^p)(y^q)(z^r)
As that would imply that p,q, and r are all even, implying
(p+1)(q+1)(r+1) = 20 to be odd, a contradiction.

Answer 87

If we assume N is perfect sq, then product of factors will be
N^((m+1)/2)
=> m=19
19 = 1*19 only
hence N = x^18 (only)
Hope you can make out where this will go.