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PriyankaPalit7
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If n is an integer, is 3^n+1 divisible by 10? Updated on: 17 Mar 2019, 04:17
Originally posted by PriyankaPalit7 on 17 Mar 2019, 03:24.
Last edited by PriyankaPalit7 on 17 Mar 2019, 04:17, edited 3 times in total.
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E
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85% (hard)

Question Stats:

35% (01:37) correct 65% (01:47) wrong based on 111 sessions

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If n is an integer, is \(3^n+1\) divisible by 10?

1) \(n=4k+2\)
2) \(n>4\)
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E
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If n is an integer, is 3^n+1 divisible by 10? 17 Mar 2019, 03:34
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If n is an integer, is \(3^n+1\) divisible by 10?

1) \(n=4k+2\)
2) \(n>4\)
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1. n=4K+2, but we dont know what k is: So not sufficient
2. n>4. So n= 5,6,7,8,9,10
so, when n=5, \(3^n+1\) is not divisible by 10, and when n=6 \(3^n+1\) is divisible by 10. So Not sufficient.

Both together: Still we dont know what k is, Hence not sufficient!!

hence IMO answer : E
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If n is an integer, is 3^n+1 divisible by 10? Updated on: 17 Mar 2019, 04:16
Originally posted by thyagi on 17 Mar 2019, 04:05.
Last edited by thyagi on 17 Mar 2019, 04:16, edited 1 time in total.
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Assuming k to be an integer will be the biggest crime done in this question....

3^n+1 will be divisible by 10 if units digit of 3^n ends with a 9...

Statement 1 - n = 4k + 2
If I assume k to be an integer, the answer would be yes. How??

If k = 1 , n = 6 and 3^6 ends with a 9
If k = 2 , n = 10 and again 3^10 ends with a 9... My answer to the main question would be yes.... But k may be or may not be an integer. Let's say k = 1/2

N = 4*1/2 + 2 = 4
3^4 + 1 won't be divisible by 10... This statement is not sufficient...

Statement 2 - same as statement 1... N can be 5,6,7,8,9...... this is also not sufficient...

Combining both statements - won't help us either... Because k can take any value... The answer must be E..

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If n is an integer, is 3^n+1 divisible by 10? 17 Mar 2019, 04:14
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1) n = 4k+2
If k >= 0, then\(3^n + 1\) will always be divisible by 10. This is because units digit of powers are 3 follow the cycle 3,9,7,1,3,9,7,1...and has a cyclicity of 4.
I.e, 3^(4k+2) has unit digit of 9. You can try with values k = 0, k = 1 and so on.
So 3^(4k+2) + 1 will always have units digit of 9 + 1 = 0.

However, in the question stem it is not mentioned that \(k >= 0\). So, if \(k < 0\), \(3^n+1\) will return fractional results - as pointed out by thyagi.
If k fractional, k = 1/2, then n = 4*1/2 + 2 = 4.
Then the units digit of \(3^n+1\) is 2 and not 0.
Hence statement 1 is insufficient.


2) \(n > 4\).
\(3^5 + 1\) has units digit of 3 + 1 = 4, \(3^6 + 1\) has units digit of 9 + 1 = 0. Two different unit's digits, one not divisible by 10 and the other is divisible by 10. Hence, insufficient.

Both statements together, we can eliminate the scenario where k < 0. So n > 4, that means k > 0. So, \(3^n + 1\) = 3^(4k+2) + 1 will have units digit 0. However, the possibility of a fractional k is still there. So even statements 1 and 2 together also do not suffice.

Answer E
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If n is an integer, is 3^n+1 divisible by 10? 06 Jul 2025, 22:39
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