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14 May 2020, 04:17
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C
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Difficulty:
95%
(hard)
Question Stats:
42% (01:55) correct
58%
(01:40)
wrong
based on 161
sessions
History
Date
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Not Attempted Yet
What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)?
A.2
B.3
C.4
D.8
E.6
A.2
B.3
C.4
D.8
E.6
14 May 2020, 04:48
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preetamsaha
What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)?
A.2
B.3
C.4
D.8
E.6
A.2
B.3
C.4
D.8
E.6
Now the difference in \((3^{2003} – 1) and (3^{2003} + 1) = (3^{2003} + 1) - (3^{2003} - 1)=2\),
Both (3^2003 – 1) and (3^2003 + 1) are even, so their HCF is 2.
We know LCM*HCF=product of the two numbers
=> \(LCM*2=(3^{2003} – 1) * (3^{2003} + 1)=(3^{2003})^2-1^2=3^{4006}-1\)
We also know that the cylicity of units digit of successive powers of 3 is 3, 9, 7, 1, 3, 9, ...
So \(3^{4006}=3^{4*1001+2}\) will have same units digit as 3^2 or 9..
So units digit of \(3^{4006}-1=9-1=8=LCM*2\)
So LCM can have units digit as X8/2 or 4, where X is even, OR Y8/2 or 9, where Y is odd. But \(LCM*2=(3^{2003} – 1) * (3^{2003} + 1)=E*E\), so LCM has to be EVEN, so 4 is the answer
C
General Discussion
yashikaaggarwal
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14 May 2020, 04:39
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Cyclicity repeats after 4 exponential in 3
Therefore 3^2000*3^3 will have unit digit if 7
3^1=3
3^2=9
3^3=27 -------(1)
Now the no. Of which we have to draw LCM is (3^2003-1)(3^2003+1)
(7-1)(7+1)
(6)(8)
Taking LCM of 6&8
That will be 24
Unit digit will be 4
Hence answer is C
Posted from my mobile device
Therefore 3^2000*3^3 will have unit digit if 7
3^1=3
3^2=9
3^3=27 -------(1)
Now the no. Of which we have to draw LCM is (3^2003-1)(3^2003+1)
(7-1)(7+1)
(6)(8)
Taking LCM of 6&8
That will be 24
Unit digit will be 4
Hence answer is C
Posted from my mobile device
