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17 Apr 2019, 23:42
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
19% (02:41) correct
81%
(02:51)
wrong
based on 63
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive two-digit integers are factors of \(2^{24}-1\)?
(A) 4
(B) 8
(C) 10
(D) 12
(E) 14
(A) 4
(B) 8
(C) 10
(D) 12
(E) 14
20 Apr 2019, 19:27
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Bunuel
\(2^{24}-1=(2^{12}-1)(2^{12}+1)=(2^6-1)(2^6+1)(2^{12}+1)=63*65*(64*64+1)=3*3*7*5*13*4097=3*3*5*7*13*17*241\)
We are looking for two digits factors..
So let us start from the biggest..
1) 241..None
2) 17...17, 17*3, 17*5.....three
3) 13.....13, 13*3, 13*5, 13*7.....four
4) 7.......7*3, 7*5, 7*3*3, ......three
5) 5.....5*3, 5*3*3.....two
Total 3+4+3+2=12
D
General Discussion
20 Apr 2019, 21:10
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(a^24-1)= (a^12-1)(a^12+1)
=(a^6-1)(a^6+1)(a^4+1)(a^8+1-a^4)
(2^24-1) = (2^6-1)(2^6+1)(2^4+1)(2^8+1-2^4)
=63*65*17*241=(3^2)*5*7*13*17*241
Now 2 digits factors of (2^24-1)=3*5, 3*7, 3*13, 3*17, (3^2)*5, (3^2)*7, 5*7, 5*13, 5*17, 7*13, 13*1, 17*1
Hence 12 values are possible.
=(a^6-1)(a^6+1)(a^4+1)(a^8+1-a^4)
(2^24-1) = (2^6-1)(2^6+1)(2^4+1)(2^8+1-2^4)
=63*65*17*241=(3^2)*5*7*13*17*241
Now 2 digits factors of (2^24-1)=3*5, 3*7, 3*13, 3*17, (3^2)*5, (3^2)*7, 5*7, 5*13, 5*17, 7*13, 13*1, 17*1
Hence 12 values are possible.
