Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Least Positive Integer.

## Least Positive Integer Problem – AIME I, 2000

Find the least positive integer n such that no matter how \(10^{n}\) is expressed as the product of any two positive integers, at least one of these two integers contains the digit 0.

- is 107
- is 8
- is 840
- cannot be determined from the given information

**Key Concepts**

Product

Least positive integer

Integers

## Check the Answer

But try the problem first…

Answer: is 8.

AIME I, 2000, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

\(10^{n}\) has factor 2 and 5

Second Hint

for n=1 \(2^{1}\)=2 \(5^{1}\)=5

for n=2 \(2^{2}\)=4 \(5^{2}=25\)

for n=3 \(2^{3}\)=8 \(5^{3}=125\)

……..

Final Step

for n=8 \(2^{8}\)=256 \(5^{8}=390625\)

here \(5^{8}\) contains the zero then n=8.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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