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# If a three digit number ‘xyz’ has 2 factors, how many factors does the

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If a three digit number ‘xyz’ has 2 factors, how many factors does the  [#permalink]

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20 Oct 2017, 10:05
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If a three digit number ‘xyz’ has 2 factors, how many factors does the 6-digit number ‘xyzxyz’ have?

A. 16
B. 18
C. 20
D. 24
E. 30

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Hasan Mahmud

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Joined: 26 Feb 2016
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If a three digit number ‘xyz’ has 2 factors, how many factors does the  [#permalink]

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20 Oct 2017, 10:40
The best way to solve this question is to find a number xyz which has 2 factors.
Only prime numbers have only 2 factors. Let the prime number be 101

If a number(N) can be prime factorized to give $$a^m*b^n*c^o$$,
the total number of factors are $$(m+1)(n+1)(o+1)$$

101101 is the 6 digit prime number, which when prime factorized gives 7*11*13*101
Therefore, the total number of factors are 2*2*2*2 = 16 factors(Option A)
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Re: If a three digit number ‘xyz’ has 2 factors, how many factors does the  [#permalink]

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20 Oct 2017, 10:54
Any six digit number xyzxyz can be re-written as: xyz*1001 or xyz * 7 * 11 * 13 (since 1001 is product of 7, 11, 13)

Now xyz is prime (its given that it has 2 factors only) so this number xyz * 7 * 11 * 13 is now the product of 4 distinct prime numbers. Hence its number of factors = 2*2*2*2 = 16

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Re: If a three digit number ‘xyz’ has 2 factors, how many factors does the  [#permalink]

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21 Oct 2017, 15:02
Mahmud6 wrote:
If a three digit number ‘xyz’ has 2 factors, how many factors does the 6-digit number ‘xyzxyz’ have?

A. 16
B. 18
C. 20
D. 24
E. 30

A six digit number xyzxyz can be re-written as: xyz*1001
or
xyz * 7 * 11 * 13

number of factors = 2*2*2*2 = 16

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Joined: 12 Oct 2017
Posts: 40
If a three digit number ‘xyz’ has 2 factors, how many factors does the  [#permalink]

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21 Oct 2017, 19:27
1
We have the formula to find the number of factors of an integer n (include 1 and n itself):

First make prime factorization of an integer n=a^p * b^q * c^r (where a,b,c are prime factors of n and p,q,r are their powers)
Then, the number of factors of n will be expressed by the formula: (p+1)(q+1)(r+1)

Here we have xyzxyz = xyz * 1001 = xyz^1 * 7^1 * 11^1 *13^1
Since xyz has 2 factors => xyz is a prime number.
=> total number of factors of xyzxyz = 2*2*2*2 = 16 factors.

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If a three digit number ‘xyz’ has 2 factors, how many factors does the &nbs [#permalink] 21 Oct 2017, 19:27
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