An Abstract Way to Celebrate the New Year!
What is the sum of the last two digits of ∣ 1 ! − 2 ! + 3 ! − 4 ! + … + 2 0 1 1 ! − 2 0 1 2 ! + 2 0 1 3 ! − 2 0 1 4 ! ∣ ?
The answer is 10.
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2 solutions
@Joshua_Ong I would name it as an absolutely fantastic way to celebrate the new year.
Thanks :)
10! itself is divisible by 100
An interesting pattern can be found.
The sum of the last 2 digits of ∣ 1 ! − 2 ! + . . . − 8 ! + 9 ! ∣ = 8 + 1 = 9 .
The sum of the last 2 digits of ∣ 1 ! − 2 ! + . . . + 9 ! − 1 0 ! ∣ = 1 + 9 = 1 0 .
Repeating this, we obtain
Sum of the last 2 digits of ∣ 1 ! − 2 ! + . . . n ! ∣ = 9 when n is odd and 1 0 when n is even.
Since 2 0 1 4 is even, the required answer is 1 0 .
= = = = ≡ = ≡ ∣ 1 ! − 2 ! + 3 ! − 4 ! + ⋯ + 2 0 1 1 ! − 2 0 1 2 ! + 2 0 1 3 ! − 2 0 1 4 ! ∣ ∣ 1 ! − 1 ! × 2 + 3 ! − 3 ! × 4 + ⋯ + 2 0 1 3 ! − 2 0 1 3 ! × 2 0 1 4 ∣ ∣ 1 ! ( 1 − 2 ) + 3 ! ( 1 − 4 ) + ⋯ + 2 0 1 3 ! ( 1 − 2 0 1 4 ) ∣ ∣ − 1 × 1 ! − 3 × 3 ! − 5 × 5 ! − 7 × 7 ! − 9 × 9 ! − 1 1 × 1 1 ! − ⋯ − 2 0 1 3 × 2 0 1 3 ! ∣ 1 × 1 ! + 3 × 3 ! + 5 × 5 ! + 7 × 7 ! + 9 × 9 ! + 1 1 × 1 1 ! + ⋯ + 2 0 1 3 × 2 0 1 3 ! 1 × 1 ! + 3 × 3 ! + 5 × 5 ! + 7 × 7 ! + 9 × 9 ! 3 3 0 1 8 1 9 1 9 (mod 100) (mod 100)
The key here is to realize that factorials greater than 10 are divisible by 100. (Proving this is left to the readers as an exercise)