Latest Elitmus Aptitude Question SOLUTION: a natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?

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(#M40008864) ELITMUS Numerical Ability question Keep an EYE

a natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?

Asked In Elitmus amrit bhagat (12 years ago)

Unsolved Read Solution (15)

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Gaurav Dhimankar  (9 years ago)

This mathematical problem can be explained well if we use a basic formula which helps to find out the the No. of Divisors or Factors of a particular Numbers it has, it generally goes like this.

If N is a Natural number, Prime Factorization of N = a^x * b^y .
Then the No. of divisors = (x+1)*(y+1)

As the Natural number has exactly 10 divisors & 10 = 5 * 2
So possible value of x = 4 & y = 1 or x = 9 .
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Example
if 512 = 2^9 , total No. of factor = (x+1) i.e (9+1) = 10
or
if 48 = 2^4*3^1 , total No. of factor = (x+1)*(y+1) i.e (4+1)*(1+1) = 10
or
if 19683 = 3^9 , total No. of factor = (x+1) i.e (9+1) = 10
or
if 162 = 2^1*3^4, total No. of factor = (x+1)*(y+1) i.e (4+1)*(1+1) = 10
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Thus Answer = Either 1 or 2 distinct prime factors.

Is this solution Helpfull? Yes (70) | No (3)  

vipul pratap singh  (11 years ago)

ans is either 1 or 2 .
example if the no is 512 . then the divisors of this no. are
1,2,4,8,16,32,64,128,256,512;

so only one prime factors are there 2;

if the no is 48 the divisors are
1,2,3,4,6,8,12,16,24,48;
so the prime factors are 2,3;

Is this solution Helpfull? Yes (49) | No (8)  

shashidhar  (11 years ago)

for eg:48 is the number with 10 divisors 1,2,3,4,6,8,12,16,24,48 and hence only 2 are possible i.e 2 and 3

Is this solution Helpfull? Yes (19) | No (9)  

Pargat Pal Singh  (12 years ago)

2 prime factors are possible only in case of natural number having exactly 10 divisors including 1 and itself.

Is this solution Helpfull? Yes (10) | No (11)  

kratika jain  (10 years ago)

it will hav 2 prime factor
as 10=5*2 thus no. will be of form (prime no.1)^4 * (prime no.2)
for eg 48= divisors(1 2 4 8 16 3 6 12 24 4 48)= 2^4*3
(it consist of two prime no.)

Is this solution Helpfull? Yes (7) | No (2)  

Sanjana Patnaik  (12 years ago)

There is only two distinct prime factors.
If No. of divisors are 10, then the expression of prime factors would be:

P^a+Q^b and then the no. of factors would be (a+1)(b+1) which 10 is possible.
If there are more than two prime factrors, 10 is not possible.

Is this solution Helpfull? Yes (6) | No (14)  

Rakesh Kumar Kohli  (12 years ago)

The number will be a multiple of 8 prime factors. Then it will be divisible by 1 and the number itself i.e it will have exactly 10 divisors.

Example:

3*7*11*13*17*19*23*29 = 646969323
This number has exactly 10 divisors including 1 and itself.

Is this solution Helpfull? Yes (3) | No (43)  

rohitashwa  (10 years ago)

as we 9 no. of factors=(p+1)(q+1).... now see if no. is n ,then divisors may be 2,3,5,or 7 or other 2 digit prime nos. . as 4 ,6,8,9 or other 2 digit nos. can be obtained by using above digits for example 4=2^2,9=3^2 nd so on.thus no n =a^p+b^q+c^r....and given no. of factor=10.so (p+1)(q+1)(r+1)....=10.as 10 can be broken down as 2*5 or 10*1, or vice versa.in each case for example 2*5=we can use at most 2 digits(2 digits m using as we r talking abt powers to a number) as either p+1=2 & q+1=5, or p+1=10 &q+1=1.as p being value of power .each of them have atmost 2 values i.e 2 when 2*5 and 1 when 10*1 (as in this case q=0 and p & q are powers so if power goes o no. goes to 1)so 1 when a^9 and 2 when a^4*b^1(for the guys who don't 9 no of factors rule no of factors of a no. n which can be represented as a^p*b^q,means no. can be written as power of prime no.s a and b is always equal to p+1*q+1.)

Is this solution Helpfull? Yes (3) | No (1)  

rajesh  (9 years ago)

as if no is 512 then
we can say that divisors are 2^9
so it will have only one prime divisor which is 2

Is this solution Helpfull? Yes (2) | No (2)  

ashes  (10 years ago)

2 IS THE ANSWER
10 FACTORS =(1+1)(4+1)
EXAMPLE 2*3*3*3*3
OR 3*5*5*5*5
.............
MANY MORE BUT ALL THE NUMBERS WILL HAVE ONLY 2 DISTINCT
PRIME FACTORS

Is this solution Helpfull? Yes (1) | No   

Abhinav khare  (9 years ago)

there can be atmost 8 prime factors
for eg 30 = 2*3*5 -3 factors along with 1 and 30 itself
similarly 210 = 2*3*5*7 - 4 factors
similarly if question says there are 10 factors including 1 an self then we can multiply upto 8 consecutive prime no.

Is this solution Helpfull? Yes (1) | No   

sravani  (9 years ago)

I think 4...

Is this solution Helpfull? Yes (1) | No (1)  

Sameer kumar  (5 years ago)

A number can be written in form of (p^a)(q^b)(r^c) where p,q,r are its prime factors.
then number of factors = (a+1)(b+1)(c+1)
in case of 10 factors, it can be :
1*10 or 2*5
i.e.
(p^9) ---> factors = (9+1) = 10 or,
(p^1)(q^4) --> factors = (1+1)(4+1) = 10
so, in first case no. of prime factors = 1 (i.e. p)
and in second case number of prime factors = 2 (i.e. p,q)
Ans : either 1 or 2 distinct prime factors.

Is this solution Helpfull? Yes (1) | No   

ROHIT YADAV  (8 years ago)

ans is 1 or 2
as --- given 10 including 1 and itself means actually it has 8 divisors
means 8=4*2
or 8= 8*1
thats why

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VIDYA LAXMI  (7 years ago)

maximum 9 prime factor can have ....

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