Exaggerated

Number Theory Level 4

∑ n = 1 2017 n 1 ! + 2 ! + 3 ! + ⋯ + 2017 ! \large\displaystyle\sum_{n=1}^{2017}{ { n }^{ 1! + 2! + 3! + \cdots+2017! } }

Find the unit digit of the above expression.


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

1 7 5 3 9

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

1 ! = 1 1! = 1

2 ! = 1 x 2 = 2 2! = 1x2 = 2

3 ! = 1 x 2 x 3 = 6 3! = 1x2x3 = 6

4 ! = 1 x 2 x 3 x 4 = 24 4! = 1x2x3x4 = 24

5 ! = 1 x 2 x 3 x 4 x 5 = 120 5! =1x2x3x4x5 = 120

Note that starting form 5 ! 5! there is always at least one 5 ∗ 2 5*2 product in the factorial. That means that if we have a 5 ∗ 2 5*2 product in the factorial, the number ends in z e r o zero . So the units digit from 5 ! 5! to 2017 ! 2017! is z e r o zero .

Now to get the units digit of the expression, we simply add 1 ! , 2 ! , 3 ! , 1!, 2!, 3!, and 4 ! 4! and that is 1 + 2 + 6 + 24 = 33 1+2+6+24=33

So the units digit of the expression is 3 \boxed{3} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

@Marvin Kalngan what you are saying is true if the question was ∑ n = 1 2017 n ! \large\displaystyle\sum_{n=1}^{2017}{ { n! }}

but in the question factorial is in the powers.

just a coincidence that answer is same for both the questions.

Ravneet Singh - 4 years, 4 months ago

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oh, I did not see the problem carefully ...

A Former Brilliant Member - 4 years, 4 months ago

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