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If a three digit positive integer ‘abc’ has 2 positive factors, how ma

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Bunuel
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If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   14 Apr 2020, 04:41
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If a three digit positive integer ‘abc’ has 2 positive factors, how many positive factors does the 6-digit number ‘abcabc’ have?

A. 16
B. 18
C. 24
D. 28
E. 30


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Re: If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   14 Apr 2020, 04:50
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Bunuel wrote:
If a three digit positive integer ‘abc’ has 2 positive factors, how many positive factors does the 6-digit number ‘abcabc’ have?

A. 16
B. 18
C. 24
D. 28
E. 30


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abc has two factors

i.e. abc must be a 3 digit prime numbers



CONCEPT: Only Prime numbers have 2 distinct factors



Critical Observation: any number of the form 'abcabc' is divisible by 1001


and 1001 = 7*11*13

i.e. \(abcabc = 7*11*13*3DigitPrime\)

If \(N = a^p*b^q*c^r...\)
where a, b, c... are distinct primes
Number of factors of \(N = (p+1)*(q+1)*(r+1)*...\)


therefore, Number of factors of abcabc = (1+1)*(1+1)*(1+1)*(1+1) = 16

Answer: Option A

To know the derivation of Property of ways of finding factors of a number is available in th video here

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Re: If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   14 Apr 2020, 07:28
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a +ve integer with 2+ve factors refers to prime integer
in this case let the abc = 103
so abcabc = 103103
factors of 103103 = 11*7*13*103 ; ( 2*2*2*2) ; 16
OPTION A

Bunuel wrote:
If a three digit positive integer ‘abc’ has 2 positive factors, how many positive factors does the 6-digit number ‘abcabc’ have?

A. 16
B. 18
C. 24
D. 28
E. 30


Are You Up For the Challenge: 700 Level Questions
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Re: If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   14 Apr 2020, 08:34
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If a three digit positive integer ‘abc’ has 2 positive factors, how many positive factors does the 6-digit number ‘abcabc’ have?

A. 16 --> correct: abc=100a+10b+c has 2 positive factors, so abc is 3 digits prime number. abcabc = 1000*(100a+10b+c)+(100a+10b+c) = (100a+10b+c)*1001=abc*7*11*13=multiple of 4 distinct prime numbers, so number of positive factors in the 6-digit number ‘abcabc’ = (1+1)^4=2^4=16
B. 18
C. 24
D. 28
E. 30
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Re: If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   07 Jun 2021, 00:23
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A three-digit positive integer ‘ abc ’ has 2 positive factors, therefore it is a prime number.

Say abc = 101.

Six-digit number: 101101 = 7 * 11 * 13 * 101 = Number of factors: (2 * 2 * 2 * 2 = 16)

Answer A
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If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink] New post   07 Jun 2021, 03:49
concept:
if N = a^p*b^q*c^r...
where a, b, c... are distinct primes
Number of factors of N = (p+1)*(q+1)*(r+1)*...

not able to come up with this idea, so this question seems really hard to solve at first
we need to draw a tree diagram for us to better understanding why this concept came as thus, I attached the file below
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prime number.jpg
prime number.jpg [ 69.14 KiB | Viewed 1981 times ]

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If a three digit positive integer ‘abc’ has 2 positive factors, how ma [#permalink]
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