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preetamsaha
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 14 May 2020, 04:17
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C
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Difficulty:

95% (hard)

Question Stats:

43% (01:49) correct 57% (01:39) wrong based on 253 sessions

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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
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C
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chetan2u
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 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
Show more


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
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 05 Jul 2025, 09:29
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The HCF of \((3^x - 1)\) and \((3^x + 1)\) for any +ve value of x would be 2. So their LCM would be \(\frac{[(3^x - 1)(3^x + 1)]}{2}\).

Now, such questions are usually based on patterns so let us find one.

\(x = 1; LCM = 4\)
\(x = 2; LCM = 40\)
\(x = 3; LCM = 364\)
\(x = 4; LCM = 40*82\), again this no. would end in 0.

voila! we found our pattern!! It is \(4,0,4,0,...\) (cyclicity of 2)

So the units digit of the LCM of \((3^{2003} - 1)\) and \((3^{2003} + 1)\) would be 4.
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yashikaaggarwal
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 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

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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 10 Jun 2025, 10:30
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yashikaaggarwal
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

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Hi,, whats the logic behind taking LCM of unit digit 6 & 8 and then taking as its unit digit as answer..??

In such case, suppose we have digits 16 and 24 ====>> LCM is 48 and last digit of LCM is 8.

But with your logic LCM of last digit of both numbers are 6 and 4 making their LCM as 12 and last digit of LCM as 2. That doesn't matches..
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 09 Jul 2025, 01:27
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Can you elaborate by HCF of two even numbers is always 2? In my opinion HCF of 4 and 12 will be 4, not 2.
chetan2u
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


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
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chetan2u
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 09 Jul 2025, 04:34
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Su1206
Can you elaborate by HCF of two even numbers is always 2? In my opinion HCF of 4 and 12 will be 4, not 2.
chetan2u
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


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
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You will have to connect this to the statement prior to it. The given two numbers have a difference of 2 between them, that isvthey are alternate numbers.
Alternate EVEN numbers will have 2 as HCF while alternate ODD will be co prime so 1 is HCF.
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What is the unit digit of the LCM of (3^2003 – 1) and (3^2003 + 1)? 28 Jul 2026, 05:47
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