How many digits are in the number (10^{100} - 9^{100})? This problem requires an understanding of logarithms and number theory to arrive at a precise answer. Let's break down the solution step-by-step.

Introduction

Knowing the number of digits in a number is often useful in various applications, from estimating large numbers to understanding the scale of phenomena in science and technology. This article will demonstrate how to determine the number of digits in the specific number provided, (10^{100} - 9^{100}), using logarithmic properties and mathematical induction.

Step 1: Estimating (9^{100})

To estimate (9^{100}), we can start with the expression (9^{100} 10^{100} - 1^{100}). Using the binomial theorem, we can expand this as follows:

[9^{100} 10^{100} - sum_{k0}^{100} binom{100}{k} 10^{100-k} - 1^k]

The first few terms of this expansion are:

(k0): (10^{100}) (k1): (-100 cdot 10^{99}) (k2): (binom{100}{2} cdot 10^{98} 4950 cdot 10^{98})

Combining these, we get:

[9^{100} approx 10^{100} - 100 cdot 10^{99} 4950 cdot 10^{98}]

Step 2: Finding (10^{100} - 9^{100})

Substituting this back into the expression (10^{100} - 9^{100}), we get:

[10^{100} - 9^{100} approx 10^{100} - left(10^{100} - 100 cdot 10^{99} 4950 cdot 10^{98}right)]

This simplifies to:

[10^{100} - 9^{100} approx 100 cdot 10^{99} - 4950 cdot 10^{98}]

Factoring out (10^{98}), we obtain:

[10^{100} - 9^{100} approx 10^{98} cdot (1000 - 4950) 10^{98} cdot (-3950)]

Step 3: Calculating the Number of Digits

To find the number of digits in (-3950 cdot 10^{98}), we use the formula:

[d lfloor log_{10} (-3950 cdot 10^{98}) rfloor 1]

Using properties of logarithms, we can break this down:

[log_{10} (-3950 cdot 10^{98}) log_{10} (-3950) log_{10} 10^{98} log_{10} 3.95 cdot 10^3 98]

Estimating (log_{10} 3.95 cdot 10^3):

[log_{10} 3.95 cdot 10^3 approx log_{10} 4 cdot 10^3 log_{10} 4 3 approx 0.6 3 3.6]

Thus,

[log_{10} (-3950 cdot 10^{98}) approx 3.6 98 101.6]

The number of digits (d) is then:

[d lfloor 101.6 rfloor 1 101 1 102]

Therefore, the number of digits in (10^{100} - 9^{100}) is (boxed{102}).

Proof by Mathematical Induction

Let (N_x) represent the number of digits in the positive integer (x). We will prove that for any positive integer (n), (10^n - 9^n) has (n) digits.

Base Cases

[n 1 implies 10^1 - 9^1 10 - 9 1text{ (1 digit)}]

[n 2 implies 10^2 - 9^2 100 - 81 19text{ (2 digits)}]

Inductive Step

Assume true for (n - 1). Then, (N_{10^{n-1} - 9^{n-1}} n - 1).

[10^n 1000cdots0 text{ (n zeros)}text{ (n 1 digits)}] [10^n - 9^n geq 10^n - 10 cdot 9^{n-1} 10 cdot 10^{n-1} - 9^{n-1}text{ (at least n digits)}]

Combining these, we have:

[10^n - 9^n geq 10 cdot 10^{n-1} - 9^{n-1}]

Thus, (N_{10^n - 9^n} geq 1 N_{10^{n-1} - 9^{n-1}} ntext{ (at least n digits)})

Since (10^n - 9^n leq 10^n), which is a number with (n 1) digits, we conclude:

[N_{10^n - 9^n} n]

Therefore, (10^{100} - 9^{100}) has (boxed{100}) digits by the given property of (n).