It just won't stop...

Algebra Level 2

$\textbf{What is the last digit of:}$

$\huge \color{#D61F06}{1!} \color{#0C6AC7}+\color{#EC7300}{2!} \color{#0C6AC7}+\color{#20A900} {3!} \color{#0C6AC7}+ \color{teal}{4!} \color{#0C6AC7}+ \color{#69047E}{5!} \color{#0C6AC7}{+ ... +} \color{hotpink}{2015!}$

$\textbf{NOTES:}$

$\text{No need to bash, in fact very simple!}$
$\text{Implied from above ... NO CALCULATORS!!!}$
$\LaTeX\text { for life!}$

The answer is 3.

1 solution

Andrew Zakharov
Jun 5, 2015

${n!} = n*(n-1)*(n-2)*...*1$

Easily notice that if any (n-i) = 10 or (n-i)*(n-k)=10, then n! ends at 0.

Factorial of any numbe greater than or equal to 10 contains 10.

Factorial of any number from 5 to 9 contains 2 and 5, and 2*5=10.

Thus 5! + 6! + ... + 2015! ends at 0.

3! + 4! = 1 2 3 + 1 2 3 4 = 1 2 3(1+4) = 1 2 3 5 = 1 3 10 = 30 ends at 0.

1! + 2! = 1 + 2 = 3.

1! + 2! + 3! + 4! + 5! + ... + 2015! ends at 3.

It just won't stop...

The answer is 3.

1 solution

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