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I don't know how to correct that part. Is this a question from a worksheet or textbook?

Find all 3-digit numbers such that the sum of the digits is 11 times less than the number.

Actually, I used Google Translate, because I did not know the grammar.

Ah okay, no worries mate.

Thank you @ParthKohli .
Now solve this.

What about 198?

oh myyy...

You don't find easy ways...

Also 0 :-)

0 is not a 3-digit number @whpalmer4 .

\[11a + 11b + 11c = 100a + 10b + c \iff -89a + b + 10c = 0\]

Yes. It is easier to find.

Yeah, let's try \(a = 1\). Then \(b = 9\) and \(c = 8\) which is exactly what @whpalmer4 said

If \(a = 2\), then \(b = 8\) but \(c > 9\).

So I guess \(198\) is the only number.

(though it will toss out 000 as a solution)

I would give you both medals if i could.

Don't worry, whpalmer has mine.

@whpalmer4 Don't worry with the 000. We can figure out trivial solutions in the end. :-D

Fun little late-night problem :-)

Holy Wolfram|Alpha!

\[17a + 17b + 17 c= 256a + 16b + c\]

That?

Yes, that's it, I think.

And the solutions can't exceed \(15\).

What's that? Mathematica?

Yes.

I'll check back in the morning to see what you guys have figured out :-)

Nah, I am pretty bad at mathematics, I'd just leave this thread and we'd on this tomorrow. =)

Work on this*

Ok. I'll try. Good night.

So, 9*10*10+1*10*8+1*9*1

Sorry, it is 8*10*10+1*10*8+1*9*1=889