A number theory problem by Dharma Teja

Number Theory Level pending

Find the unit digit of 1 ! + 2 ! + 3 ! + … + 2014 ! 1!+2!+3!+\ldots+2014!

Note: n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) … 3 ⋅ 2 ⋅ 1 n! = n\cdot(n-1)\cdot(n-2)\ldots3\cdot2\cdot1


The answer is 3.

Dharma Teja by Dharma Teja , 20, India

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1 solution

Parveen Soni
Nov 30, 2014

n! has unit digit zero if n>4. So n! where n<=4 will effect the unit digit.
unit digit of 1!=1 and 2!=2 and 3!=6 and 4!=4 so sum=13 so unit digit will be 3

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