Find the last digit of ( 1 ! + 2 ! + 3 ! + ⋯ 3 3 ! ) 3 3 .
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Beautiful.
If you want last two digits then it will be more challenging
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Yes, it is. I'm getting an answer of
1 3 3 3 ≡ ( − 3 ) 1 1 ( m o d 1 0 0 ) ≡ 5 3 ( m o d 1 0 0 ) .
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Yes I also got this
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@Kushal Bose – Since it was your idea, did you want to post the mod 100 version of this problem or shall I?
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@Brian Charlesworth – U can do it I have no problem
Nice solution. We know that 3 3 3 ≡ 3 ( m o d 1 0 ) because 3 2 ≡ − 1 m o d 1 0 , and so 3 3 3 = 3 2 ∗ 1 6 + 1 ≡ ( ( − 1 ) 1 6 × 3 ) ( m o d 1 0 ) ≡ 3 ( m o d 1 0 ) .
Thank you all.@Brian Charlesworth Sir can you tell me how did you compute 1 3 3 3 ≡ 5 3 ( m o d 1 0 0 )
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This is probably longer than necessary, but here is my computation.
Since the last two digits of 1 0 ! are 0 0 we only need to consider
( 1 ! + 2 ! + 3 ! + 4 ! + 5 ! + 6 ! + 7 ! + 8 ! + 9 ! ) ( m o d 1 0 0 ) ≡
( 1 + 2 + 6 + 2 4 + 2 0 + 2 0 + 4 0 + 2 0 + 8 0 ) ( m o d 1 0 0 ) ≡
2 1 3 ( m o d 1 0 0 ) ≡ 1 3 ( m o d 1 3 ) .
Now 1 3 3 = 1 3 × 1 6 9 ≡ 1 3 × 6 9 ( m o d 1 0 0 ) ≡
1 3 ( 7 0 − 1 ) ( m o d 1 0 0 ) ≡ ( 9 1 0 − 1 3 ) ( m o d 1 0 0 ) ≡
− 3 ( m o d 1 3 ) . So now we have that
1 3 3 3 = 1 3 3 ∗ 1 1 ≡ ( − 3 ) 1 1 ( m o d 1 0 0 ) ≡ − 3 ∗ ( 3 5 ) 2 ( m o d 1 0 0 ) ≡
− 3 ∗ 4 3 2 ( m o d 1 0 0 ) ≡ − 3 ∗ ( 4 0 + 3 ) 2 ( m o d 1 0 0 ) ≡
− 3 ∗ ( 1 6 0 0 + 2 4 0 + 9 ) ( m o d 1 0 0 ) ≡ − 3 ∗ 4 9 ( m o d 1 0 0 ) ≡
3 ∗ 5 1 ( m o d 1 0 0 ) ≡ 5 3 ( m o d 1 0 0 ) .
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Solution:
Dividing the expression by 10 will help us to find the last digit,
( 1 ! + 2 ! + 3 ! + 4 ! ⋯ 3 3 ! ) ( 1 ! + 2 ! + 3 ! + 4 ! ⋯ 3 3 ! ) 3 3 3 3 3 ≡ 1 + 2 + 4 + 6 + 0 + 0 . . . . . + 0 ( m o d 1 0 ) ( 1 0 ∣ n ! ≥ 5 ) ≡ 3 3 3 ( m o d 1 0 ) ≡ 3 ( m o d 1 0 )
Last digit of ( 1 ! + 2 ! + 3 ! ⋯ 3 3 ! ) 3 3 = 3