## How many integers $0 \le n \lt 10^{18}$ have the property that the sum of the digits of $n$ equals the sum of digits of $137n$?

### To answer this problem, let’s begin by giving a general overview of the problem and then going further into detail.

1. **General Overview & Initial Observations:**

First, consider the basic rule of arithmetic: if $a \equiv b$ (mod m), then $c*a \equiv c*b$ (mod m).

But you’ll notice that the question has to do with digit-sums, specifically mod 9 digit sums (since such sums result in an integer 0-8 that represents the remainder when the original number is divided by 9). However, our initial observation does not hold for additions and subtractions mod 9, so we need a different approach.

2. **Specific Observations:**

This property is true when n is a multiple of 3. Why? Because if a number n is divisible by 3, it remains unchanged or reduces (mod 9), while its scaled version, in this case 137n, which is also a multiple of 3, will have the addition of the digits still divisible by 3. Therefore, their digit-sums which are both multiples of 3 should be equal in order to satisfy the requirement given in the problem.

Now, why only multiples of 3? Notice that 137 is also a multiple of 3 (since the sum of its digits 1+3+7 = 11, which is not a multiple of 3). This makes all 137n multiples of 3.

However, not all multiples of 3 have this property. For that, 137n should not exceed 10^(18) because if the sum of the digits of n is equal to the sum of the digits of 137n, then the number of digits in n and 137n should be equal. If the number of digits in 137n gets higher than the number of digits in n, then this property does not hold, as the increase in the number of digits will make the sum of the digits in 137n larger than that of n.

3. **Conclusion:**

Considering the number of 3’s in the given interval, we have:

For $n\le10^{18}$, there are $\frac{10^{18}}{3}=3.33*[10^{17}]$ integers that are divisible by 3. However, we must subtract the numbers in which 137n would make the number of digits expand beyond 18 digits (i.e. where n >= $10^{15}$). Therefore, there are $10^{15}/3$ such numbers.

So, putting it together, we have $3.33*[10^{17}] – 10^{15}/3$ = $3.33*[10^{17}] – 0.33*[10^{15}]$ = $3 * 10^{17}$ numbers in the given range that satisfy the stated condition.

Therefore, there are $3 * 10^{17}$ integers $0 \le n < 10^{18}$ that have the property that the sum of the digits of $n$ equals the sum of digits of $137n$. This seems very counter-intuitive because it's a unique property of modular arithmetic and properties of numbers. It shows the beauty of basic number theory and its use in problem-solving.

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