/* 
 * Solution to Project Euler problem 63
 * Copyright (c) Project Nayuki. All rights reserved.
 * 
 * https://www.nayuki.io/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */

import java.math.BigInteger;


public final class p063 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p063().run());
	}
	
	
	/* 
	 * Let's examine n^k for different values of n and k and see which
	 * choices cannot possibly work (i.e. not being exactly k digits long).
	 * 
	 * When n = 10, for each k, n^k has exactly k+1 digits, so these are excluded.
	 * By extension, when n > 10, for each k, n^k has at least k+1 digits, so these are excluded.
	 * Thus we should only consider 1 <= n <= 9.
	 * 
	 * When n = 9, k = 22, then n^k has 21 digits which is insufficient.
	 * Extending this, when n = 9 and k > 22, n^k has fewer than k digits.
	 * Furthermore, when n < 9, n^k will have start to have
	 * fewer than k digits at some value of k with k < 22.
	 * Therefore we should only consider 1 <= k <= 21.
	 * 
	 * We handle the rest of the testing by brute force.
	 */
	public String run() {
		int count = 0;
		for (int n = 1; n <= 9; n++) {
			for (int k = 1; k <= 21; k++) {
				if (BigInteger.valueOf(n).pow(k).toString().length() == k)
					count++;
			}
		}
		return Integer.toString(count);
	}
	
}