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14 Apr 2020, 04:41
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (02:09) correct
34%
(02:04)
wrong
based on 176
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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
Are You Up For the Challenge: 700 Level Questions
A. 16
B. 18
C. 24
D. 28
E. 30
Are You Up For the Challenge: 700 Level Questions
14 Apr 2020, 04:50
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Bunuel
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 \(N = (p+1)*(q+1)*(r+1)*...\)
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
General Discussion
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
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
