The sums are products

for a two digit number, say 10a + b = 3(a+b)

implies, 7a = 2b

first easy grab is 27.

we could have posed another question saying a number is eight times the sum of its digits for 72

We first note that for any $n-$ digit number, that the maximum digit sum is $9n$ and so three times the digit sum would have a maximum of $27n$ .

However we clearly see for $n > 2$ that $27n$ does not have $n$ digits. Hence, the number must have $n\leq2$ digits.

It is clear to see that there exists no such $1$ -digit number, meaning our peculiar number must have two digits.

As the number is three times the digit sum, it must also be a multiple of three, hence the digit sum must also be a multiple of three. This then means that the number and hence the digit sum must be a multiple of nine, as this is true for all two-digit numbers that are multiples of nine.

As we know the digit sum is a multiple of nine, we then see that the number itself must be a multiple of $27$ , leaving only $27,$ and $81$

We simply check to see that $27$ is the answer.

Seems like we have been woke up middle of the sea and were told to catch a fish by reading one descriptive sentence. But there are lots of fish in the sea, aren't there? So let's narrow down the neighborhood of that particular fish.

First of all, we have to know how many digits that integer has. Because it is very easy for a two digit integer: 3(x+y)=10x+y, for three digits it's 3(x+y+z)=100x+10y+z, and so on.

We can see, for a two or one digit integer, only one proper condition is sufficient. For a three digit integer two proper conditions are needed and so on.

For a single digit integer, it is very easy to verify that none of 3,6 or 9 is the answer, moreover, the question wouldn't be posted here if it was a single digit integer! We have been given only one condition. So it's a two digit integer.

Now, 3(x+y)=10x+y => 7x=2y => x=(2/7)y

Both x and y are integers and y is such that it is divisible by 7.

So, easily we can find out that the integer is 27, this is the answer.

Now, is that the only answer? The question tells so. Secondly, let's check for the other values for y. Remember, for (10x+y) to be a two digit integer, the value of x must consist single digit.

Then, let's be rebels and start thinking outside the box (a very popular concept of present age), id est, the question! What if, despite what the question implies, there are such integers of three or more digits?

Let's explore the galaxy of such integers, shall we? For this mission, I shall only give two hints: 1) The largest two digit integer divisible by 3 is 99, for three digits, it's 999, and so on. 2) 34 is the first integer which will produce a three digit integer when multiplied by 3. For four digit, the integer is 334.

How I did it was, I thought:

1. It can't be three digit number, nor can it be a single digit number (in the question it says 'sum of its digits'.
2. It says three times the sum of its digits. So, the number MUST be multiples of three.
3. Just list out multiples of three. Then you find out 27 is the correct answer.

From 1 to 100, the range of the sum of the digits of the numbers is between 1 to 18, inclusive.

Since the positive integer has to be 3 times the sum of its digits, lets observe a multiplication table of 3, from 1 to 18 which essentially is a list of the possible values of the positive integer.

Select the number of the product of the multiplication of 3, such that the sum of the product's digits matches the multiplier of 3. Eg: ... 3x7=21 [sum of digits of product=3]
3x8=24 [6]
3x9=27 [9]
...