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01 Oct 2018, 04:55
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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:
35%
(medium)
Question Stats:
67% (01:54) correct
33%
(02:09)
wrong
based on 283
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If x is a positive integer, does the remainder, when \((7^x + 1)\) is divided by 100, have 0 as the units’ digit?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 5
(1) x = 4n + 2, where n is a positive integer.
(2) x > 5
01 Oct 2018, 05:11
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Bunuel
CONCEPT: Remainder when any number is divided by hundred are the last two digits of the number e.g. 137 divided by 100 leaves 37 remainder and likewise 923 divided by 100 leaves 23 remainder etc
i.e. we only need to know if the unit digit of \((7^x + 1)\) is Zero or NOT
i.e. we need to know if \(7^x\) has unit digit 9 or not which is possible when \(x = 4a+2\) because cyclicity of 7 is 4 and \(7^2\) and \(7^6\) and \(7^10\) etc have unit digit 9
Statement 1: x = 4n + 2, where n is a positive integer.
i.e. \(7^x\) will always have unit digit 9 hence
SUFFICIENT
\(Statement 2: x > 5\)
x may be 6 or 7 etc. hence unit digit of \(7^x\) may or may not be 9 hence
NOT SUFFICIENT
Answer: Option A
Archit3110
Major Poster
16 Apr 2019, 19:16
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Bunuel
\((7^x + 1)\)
#1
x=4n+2
value of x will be even always ; x=2,6,10 at n=0,1,2
by cyclicity of 7 we know
7^1=7
7^2=9
7^3=3
7^4=1
so we see x= 2,6,10.. we get unit digit as 9 so 9+1 ;10 unit digit is always 0
#2
x > 5
not sufficient as value of unit digit of 7^x will vary
IMO A
