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Number Theory Level 4

Find the largest value of integer $n$ such that $10^n$ divides $101^{100} - 1$ .

The answer is 4.

2 solutions

Guillermo Templado
Feb 22, 2016

$101^{100} - 1 = (100 + 1)^{100} - 1 = \sum_{n=0}^{100} {100 \choose n} 100^{n} - 1 = \sum_{n=1}^{100} {100 \choose n} 100^{n} \Rightarrow$ the greatest power of 10 which perfectly divides to $101^{100} - 1$ is 4 . Look at ( ${100 \choose 1} \cdot 100^{1}$ and ${100 \choose 2} \cdot 100^{2}$ )

Your solution does show that $10^4\mid 101^{100}-1$ but how do you show that $10^n\not\mid 101^{100}-1~\forall~n\gt 4$ ?

Prasun Biswas - 5 years, 3 months ago

Because if you sum $a \cdot 100^3 + b \cdot 100^4 +.... + c \cdot 100^{100} \text{ which they have at least six 0's at the end, }$ to ${100 \choose 1} \cdot 100^{1}$ + ${100 \choose 2} \cdot 100^{2} = 49510000$ it won't affect to the last six figures(digits)

Guillermo Templado - 5 years, 3 months ago

You should include this part in your solution.

Prasun Biswas - 5 years, 3 months ago

@Prasun Biswas – good, it's just included, isn't it?

Guillermo Templado - 5 years, 3 months ago

William Isoroku
Feb 22, 2016

Use binomial theorem

I did it that way too.

Rishik Jain - 5 years, 3 months ago

Remember your identities

The answer is 4.

2 solutions

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