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![3^7[(3^−4+3^−5)+(3^−5+3^−6)+(3^−6+3^−7)–3^−7] is what multiple of 17?](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "3^7[(3^−4+3^−5)+(3^−5+3^−6)+(3^−6+3^−7)–3^−7] is what multiple of 17?"){kind=icon}
12 Mar 2020, 23:39
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:14) correct
25%
(02:34)
wrong
based on 473
sessions
History
Date
Time
Result
Not Attempted Yet
\(3^7[\)(\(3^{−4}\)+\(3^{−5}\))+(\(3^{−5}\)+\(3^{−6}\))+(\(3^{−6}\)+\(3^{−7}\)) –\(3^{−7}\) ] is what multiple of 17?
A) 1
B) 2
C) 3
D) 4
E) 11
A) 1
B) 2
C) 3
D) 4
E) 11
13 Mar 2020, 05:36
Kudos
Bookmarks
rheam25
Take: \(3^7[\)(\(3^{−4}\)+\(3^{−5}\))+(\(3^{−5}\)+\(3^{−6}\))+(\(3^{−6}\)+\(3^{−7}\)) –\(3^{−7}\) ]
Expand to get: \(3^3 + 3^2 + 3^2 + 3^1 + 3^1 + 1 - 1\)
Evaluate: \(27 + 9 + 9 + 3 + 3 + 1 - 1\)
Evaluate: \(51\)
Since \(51 = (3)(17)\), the correct answer is C
Cheers,
Brent
General Discussion
29 Nov 2020, 18:38
Kudos
Bookmarks
We can simplify the equation.
3^7 [3^-4 + 2(3^-5) + 2(3^-6) ]
Note that +3^-7 and -3^-7 are cancelled in the above simplification.
Take 3^-4 as common multiple.
3^7 [3^-4 {1 + 2(3^-1) + 2(3^-2)}]
3^3 ( 1 + 2/3 + 2/9)
We get 3^3 since 3^7 (3^-4) = 3^3
3^3 (17/9)
3 (17)
Thus the multiple is 3.
Posted from my mobile device
3^7 [3^-4 + 2(3^-5) + 2(3^-6) ]
Note that +3^-7 and -3^-7 are cancelled in the above simplification.
Take 3^-4 as common multiple.
3^7 [3^-4 {1 + 2(3^-1) + 2(3^-2)}]
3^3 ( 1 + 2/3 + 2/9)
We get 3^3 since 3^7 (3^-4) = 3^3
3^3 (17/9)
3 (17)
Thus the multiple is 3.
Posted from my mobile device
