lets play with factorials
P = 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + 1 2 ⋅ 1 2 !
Find the remainder when P + 4 is divided by 13.
The answer is 3.
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2 solutions
P = 1 × 1 ! + 2 × 2 ! + 3 × 3 ! + ⋯ + + 1 2 × 1 2 ! = r = 1 ∑ 1 2 r ⋅ r ! = r = 1 ∑ 1 2 ( r + 1 − 1 ) ⋅ r ! = r = 1 ∑ 1 2 ( r + 1 ) r ! − r ! = r = 1 ∑ 1 2 ( r + 1 ) ! − r ! = 1 3 ! − 1 ! ≡ − 1 ( m o d 1 3 ) Sum telescopes
∴ P + 4 ≡ − 1 + 4 = 3 ( m o d 1 3 )
Given 1 ⋅ 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + 1 2 ⋅ 1 2 ! .
Now, formulating the above statement it can be written as P = ( 2 − 1 ) . 1 ! + ( 3 − 1 ) . 2 ! + ( 4 − 1 ) . 3 ! + . . . . . . . . . . . . . + ( 1 3 − 1 ) . 1 2 ! P = 2 . 1 ! − 1 ! + 3 . 2 ! − 2 ! + 4 . 3 ! − 3 ! + . . . . . . . . . . . . + 1 3 . 1 2 ! − 1 2 ! Now we know that ( n + 1 ) ! = ( n + 1 ) . n ! P = 2 ! − 1 ! + 3 ! − 2 ! + 4 ! − 3 ! + . . . . . . . . . . + 1 3 ! − 1 2 ! ∴ P = 1 3 ! − 1 ! ⇒ P + 4 = 1 3 ! − 1 + 4 = 1 3 ! + 3 .
Now we see that 1 3 ! is divisible by 1 3 . So, remainder = 3