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

- klimenkov

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

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- anonymous

that are more than the sum of its digits in 11 <===that part of the question is confusing. Did you write the question properly, because it doesn't make sense to me?

- anonymous

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

- klimenkov

This is my translation. You have to find all 3-digit numbers, the sum of digits of which is 11 times less than the number. Can you write this correctly in English?

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## More answers

- ParthKohli

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

- klimenkov

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

- anonymous

Ah okay, no worries mate.

- klimenkov

Thank you @ParthKohli .
Now solve this.

- ParthKohli

So translating your question into an equation:\[\dfrac{100a + 10b + c}{11} = a + b + c\]Where the digits are \(a,b,c\).

- ParthKohli

Another way to think is to see the multiples of \(11\), because we are sure that our three-digit number is a multiple of \(11\):
110 - no.
121 - no.
132 - no.
143 - no.
154 - no.
165 - no.
Too long.

- whpalmer4

What about 198?

- ParthKohli

oh myyy...

- klimenkov

You don't find easy ways...

- whpalmer4

Also 0 :-)

- klimenkov

0 is not a 3-digit number @whpalmer4 .

- ParthKohli

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

- klimenkov

Yes. It is easier to find.

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- whpalmer4

for (i = 0; i < 10; i++)
for (j = 0; j < 10; j++)
for (k = 0; k < 10; k++)
if ((i+j+k)*11 == (i*100+j*10+k)) then printf("%d%d%d is solution", i, j, k)
works too

- whpalmer4

(though it will toss out 000 as a solution)

- klimenkov

I would give you both medals if i could.

- ParthKohli

Don't worry, whpalmer has mine.

- ParthKohli

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

- whpalmer4

WolframAlpha writes the solution as
\[b=9a + 10n, c = 8a-n, n \in Z\]
which makes it clear that 198 is the only solution if a b and c all have to be single digits

- whpalmer4

Fun little late-night problem :-)

- ParthKohli

Holy Wolfram|Alpha!

- whpalmer4

So, here's my parting shot — what about a similar problem, except in base 16, and the multiple is 17 * the sum of the digits? :-)

- ParthKohli

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

- ParthKohli

That?

- whpalmer4

Yes, that's it, I think.

- ParthKohli

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

- klimenkov

Do[If[11*(IntegerDigits[i][[1]] + IntegerDigits[i][[2]] + IntegerDigits[i][[3]]) == i, Print[i]], {i, 100, 999}]
198

- whpalmer4

Here's a question: what effect does the multiplier have on the number of solutions?
in base 16, a multiplier of 11 gets all sorts of solutions. a multiplier of 17 gets 1, a multiplier of 19 gets a small handful.

- ParthKohli

What's that? Mathematica?

- klimenkov

Yes.

- ParthKohli

So does that mean this:
If we are looking at base \(n\), and the multiplier is \(n + 1\), then there's one solution.

- whpalmer4

For no good reason, I thought that a prime number as the multiplier would lead to few solutions, and a composite number many, but that doesn't seem to be true

- whpalmer4

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

- ParthKohli

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

- ParthKohli

Work on this*

- whpalmer4

I'm confident that our friend @klimenkov will have written a program to analyze it before long :-)
With that, I'm out of here for the night!

- klimenkov

Ok. I'll try. Good night.

- anonymous

11(a+b+c)<100a+10b+c
b+10c<89a
Always true when a>=2
When a=1
always true for 0<=c<=7 and 0<=b<=9
always true when c=8 and 0<=b<=8

- anonymous

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

- whpalmer4

@LOOSEr the problem is that 9*10*10+1*10*8+1*9*1 = 989 which is not divisible by 11, and 9+8+9 = 26 * 11 = 286 so it fails to meet the requirements. Because the maximum sum of the digits is 9+9+9 = 27, and 27*11 = 297, any solution to this problem must be <= 297...

- anonymous

No.... what I mean is
Number of 3-digit numbers such that the sum of the digits is 11 times less than the number = 989

- anonymous

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

- whpalmer4

Ah, okay, I agree with that, as I find 110 numbers where the sum of the digits * 11 is >= the number, and we aren't counting 0, so there are 999 candidates in all. Sorry for misunderstanding your point earlier.

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