# 2001 IMO Shortlist Problems/N1

## Problem

Prove that there is no positive integer such that, for , the leftmost digit (in decimal notation) of equals .

## Solution

Suppose that there is such a number . Let be the number of digits of , and let be the number of digits of . Then we have:

and

Combining these inequalities via multiplication we get:

From this inequality, it can be seen that has at least and at most digits. However, if it has digits then the first digit is less than , which contradicts with how is defined. Thus, must have digits. Expressing this as an inequality, we get:

Combining these values with the earlier bounds for via division, we get:

Suppose that . Then is between and . Thus, one of the values must be . Call that value . This means that the digits of are just the digits of , followed by zeroes. Thus the first digit of the two numbers are equal, which contradicts with how is defined. This means that:

Let be the number of digits of . Then we have:

Combining these with the bounds for , and via multiplication we get:

So has digits, but is smaller than the smallest digit number with the first digit 4, which means that its first digit can't be 4. This contradicts with how is defined, thus no such value of exists.

~Circling