Simple Simplification

Number Theory Level 4

∑ k = 1 14 k × k ! \sum _{ k=1 }^{ 14 }{ k\times k! }

Find the last four digits of the expression.

For example, if the answer is 12345, enter 2345.


The answer is 7999.

Tejas Ramdas by Tejas Ramdas , 21, USA

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1 solution

Tejas Ramdas
Nov 22, 2015

k × k ! = ( k × k ! ) + k ! − k ! = ( k + 1 ) ( k ! ) − k ! = ( k + 1 ) ! − k ! k\times k! = (k\times k!) + k! - k! = (k+1)(k!) - k! = (k+1)! - k! Thus, ∑ k = 1 14 k × k ! \sum _{ k=1 }^{ 14 }{ k\times k! } can also be written as ∑ k = 1 14 ( k + 1 ) ! − k ! \sum _{ k=1 }^{ 14 }{ (k+1)! - k!} This expression evaluates to 2 ! − 1 ! + 3 ! − 2 ! + . . . + 14 ! − 13 ! + 15 ! − 14 ! 2! - 1! + 3! - 2! + ... +14! - 13! + 15! - 14! which leaves us with 15 ! − 1 ! 15! - 1! the last four digits of which are 7999 \boxed { 7999 }

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While I agree with the transformation of expression to 15 ! − 1 15!-1 , your question leaves a room on how to achieve 7999. I will continue it from there. As 15 ! 15! may be arrange to

15 ! = ( 2 × 5 ) × ( 3 × 4 × 6 × 7 × 8 × 9 ) × 10 × ( 11 × 13 × 14 ) × ( 12 × 15 ) = 1 0 3 × 36288 × 36036 15! = (2\times5)\times(3\times4\times6\times7\times8\times9)\times10\times(11\times13\times14)\times(12\times15)= 10^3\times36288\times36036 To which you can find that 15 ! = 1 0 3 × [ ( 1296 × 1 0 10 ) + ( 324 × 3 6 5 ) + ( 288 × 36 ) ] 15! = 10^3\times[(1296\times10^{10})+(324\times36^5)+(288\times36)] or in the form of 15 ! = ζ ( 1 0 8 ) + 10 , 368 , 000 15! = \zeta(10^8)+10,368,000 , therefore, the last four digits of this sum is 15 ! − 1 ( m o d 10000 ) ≡ [ 1 0 4 ( ζ ( 1 0 4 ) + 1036 ) + 7999 ] ( m o d 10000 ) = 7999 15!-1\pmod{10000} \equiv [10^4(\zeta(10^4)+1036)+7999]\pmod{10000} = 7999

Kay Xspre - 5 years, 6 months ago

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