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20 Apr 2019, 21:51
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Difficulty:
75%
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Question Stats:
56% (02:10) correct
44%
(02:31)
wrong
based on 175
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Date
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If n is a positive integer greater than 2, then \(3^{4n} - 4^{3n}\) is always divisible by
A. 13
B. 17
C. 19
D. 23
E. 7
A. 13
B. 17
C. 19
D. 23
E. 7
anshkool4u
Current Student
Joined: 18 Oct 2018
Last visit: 22 Oct 2022
Posts: 91
Given Kudos: 23
Location: India
Concentration: Finance, Operations
Schools: IIMA PGPX'23 (A)
GMAT 1: 710 Q50 V36
GPA: 4
WE:Business Development (Retail Banking)
20 Apr 2019, 22:39
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By rule,
A^n- B^n is always divisible by A-B.
So,here 3^4n = 81^n
4^3n = 64^n
Hence ,81^n- 64^n is always divisible by 81-64= 17
Hence the answer
Posted from my mobile device
A^n- B^n is always divisible by A-B.
So,here 3^4n = 81^n
4^3n = 64^n
Hence ,81^n- 64^n is always divisible by 81-64= 17
Hence the answer
Posted from my mobile device
General Discussion
24 Jan 2025, 09:35
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