Question

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

Answer 1

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?

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

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

Answer 6

Ah okay, no worries mate.

Answer 7

Thank you @ParthKohli .
Now solve this.

Answer 8

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

Answer 9

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.

Answer 10

What about 198?

Answer 11

oh myyy...

Answer 12

You don't find easy ways...

Answer 13

Also 0 :-)

Answer 14

0 is not a 3-digit number @whpalmer4 .

Answer 15

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

Answer 16

Yes. It is easier to find.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

(though it will toss out 000 as a solution)

Answer 22

I would give you both medals if i could.

Answer 23

Don't worry, whpalmer has mine.

Answer 24

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

Answer 25

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

Answer 26

Fun little late-night problem :-)

Answer 27

Holy Wolfram|Alpha!

Answer 28

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? :-)

Answer 29

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

Answer 30

That?

Answer 31

Yes, that's it, I think.

Answer 32

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

Answer 33

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

198

Answer 34

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.

Answer 35

What's that? Mathematica?

Answer 36

Yes.

Answer 37

So does that mean this:

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

Work on this*

Answer 42

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!

Answer 43

Ok. I'll try. Good night.

Answer 44

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

Answer 45

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

Answer 46

@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...

Answer 47

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

Answer 48

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

Answer 49

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.