A number theory problem by Pi Han Goh

Number Theory Level 2

2 200 × 5 500 = 49 … … 5 0000000 … 0 ⏟ all 0’s \large 2^{200} \times 5^{500} = 49\ldots \ldots 5\underbrace{0000000\ldots 0}_{\text{all 0's}}

The last n n digits of the product 2 200 × 5 500 2^{200} \times 5^{500} consists of all 0's. What is the maximum value of n n ?

200 500 300 400

Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

Rewrite the product as 1 0 200 × 5 300 10^{200} \times 5^{300} . Trailing zeros can occur only for powers of 10 10 .

So maximum number of zeros is 200 200 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Anandmay Patel
Aug 3, 2016

I think it is the 'only' value of n(not maximum).

0 Helpful 0 Interesting 0 Brilliant 0 Confused

n n can take values from 1, 2, 3, ... , 200

Pi Han Goh - 4 years, 10 months ago

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