Really Really Troublesome

Number Theory Level pending

Find the unit's place of the following expression:

1 1 1 ! + 2 ! + . . . + 10 ! + 1 2 1 ! + 2 ! + . . . + 10 ! + ⋯ + 1 9 1 ! + 2 ! + . . . + 10 ! + 2 0 1 ! + 2 ! + . . . + 10 ! 11^{1! + 2! + ... + 10!} + 12^{1! + 2! + ... + 10!} + \cdots + 19^{1! + 2! + ... + 10!} + 20^{1! + 2! + ... + 10!}

0 1 4 Too long to determine

Tanay Gaurav by tanay gaurav , 24, India

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1 solution

Tanay Gaurav
Dec 12, 2015

This expression is equivalent to 2^(1! + 2! +...) + ... + 9^(1! + 2! +...) REM[(1! + 2! + ... +10!)/4] = REM[(1 + 2 + 6)/5] = 1 Now for 2 unit digit=2, for 3= 3, for 4=4(The whole expression is odd because 1! is odd and rest even), for 5=5, for 6=6, for 7=7, for 8=8, for 9=9[please refer to cyclicity of unit digit if facing any difficulty or feel free to contact me].So unit digit =1+2+3+....9=5(answer)Sorry for the inconvenience that the correct answer is shown 4.

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