Consistent last digit?
The last digit of
3
3
is 7.
The last digit of
3
3
3
is 7.
The last digit of
3
3
3
3
is 7.
The 3 statements above are all true. Is it true that for all integers n ≥ 2 , n number of 3’s 3 3 3 3 ⋅ ⋅ 3 has a last digit of 7?
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2 solutions
just write it in a piece of paper then take a picture and submit that picture as a solution.
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I have edited my solution , is it fine now?
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Yup! It's good! Upvoted! Do you want to create a "last 2-digit version" of this problem?
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@Pi Han Goh – Sir , I think this would be true that for n greater than equal to 3 , last two digits would be 87.
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@Ankit Kumar Jain – sounds about right. Do you know how to prove it?
What about the last 3 digits? and the last 4 digits? Have you heard of Graham's number?
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@Pi Han Goh – Sir , I came to a conclusion that for all n ≥ k , the last k digits = last k digits of 3 l a s t ( k − 1 ) d i g i t s .
I couldn't phrase that well so I am giving an example like ,
last digit = 7
last 2 digits = last two digits of 3 7 ⇒ 8 7
last 3 digits = last 3 digits of 3 8 7 ⇒ 3 8 7
last 4 digits = last 4 digits of 3 3 8 7 ⇒ 5 3 8 7 and so on.
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@Ankit Kumar Jain – Go ahead and post that as a problem then!
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@Pi Han Goh – Am I right enough?
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@Ankit Kumar Jain – Not quite. You're very close. That depends on how you phrase your question though.
Go ahead and post your problem, if it's wrong, somebody will report it with a proper explanation, if it's right, then good for you, then you can post a solution too! Either way, you learn!
@Pi Han Goh – And as for the Graham ' s number , sir I didn't know about it.
But I will go for the wiki page , but sir can you please brief me about it so that I can understand it better?
Finding the last digit of power, we have to make a division by 1 0 .. 3 3 3 = 3 2 7 ≡ 7 ( m o d 1 0 ) → 3 3 3 3 = 3 3 2 7 ≡ 3 7 ≡ ( 3 2 ) 2 ∗ 3 3 ≡ 3 3 ≡ 7 ( m o d 1 0 ) … → 3 3 3 3 3 3 … 3 ≡ 7 ( m o d 1 0 ) So, the last digit always remain the same.. This is because the increasing power of 3 contains \color\red{3^{27}} every time.. And it is clear that, 3 2 7 ≡ 7 ( m o d 1 0 )
What is the computation order?
Consider , 3 2 k + 1 .
3 2 k + 1 ≡ 9 k ⋅ 3 ≡ 3 ( m o d 4 ) ⇒ 3 odd ≡ 3 ( m o d 4 ) .
∵ (n - 2) number of 3’s 3 3 3 3 ⋅ ⋅ 3 ≡ 1 ( m o d 2 ) .
⇒ (n - 1) number of 3’s 3 3 3 3 ⋅ ⋅ 3 = 3 (n - 2) number of 3’s 3 3 3 3 ⋅ ⋅ 3 ≡ 3 2 k + 1 ≡ 3 ( m o d 4 )
⇒ n number of 3’s 3 3 3 3 ⋅ ⋅ 3 = 3 (n - 1) number of 3’s 3 3 3 3 ⋅ ⋅ 3 ≡ 3 4 k + 3 .
∵ 3 4 ≡ 1 ( m o d 1 0 ) .
∴ 3 4 k + 3 ≡ 3 3 ≡ 7 ( m o d 1 0 )