## klimenkov 2 years ago Find all 3-digit numbers such that the sum of the digits is 11 times less than the number.

1. Azteck

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?

2. Azteck

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

3. 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?

4. ParthKohli

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

5. klimenkov

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

6. Azteck

Ah okay, no worries mate.

7. klimenkov

Thank you @ParthKohli . Now solve this.

8. ParthKohli

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

9. 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.

10. whpalmer4

11. ParthKohli

oh myyy...

12. klimenkov

You don't find easy ways...

13. whpalmer4

Also 0 :-)

14. klimenkov

0 is not a 3-digit number @whpalmer4 .

15. ParthKohli

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

16. klimenkov

Yes. It is easier to find.

17. ParthKohli

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

18. ParthKohli

If $$a = 2$$, then $$b = 8$$ but $$c > 9$$.

19. ParthKohli

So I guess $$198$$ is the only number.

20. 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

21. whpalmer4

(though it will toss out 000 as a solution)

22. klimenkov

I would give you both medals if i could.

23. ParthKohli

Don't worry, whpalmer has mine.

24. ParthKohli

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

25. 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

26. whpalmer4

Fun little late-night problem :-)

27. ParthKohli

Holy Wolfram|Alpha!

28. 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? :-)

29. ParthKohli

$17a + 17b + 17 c= 256a + 16b + c$

30. ParthKohli

That?

31. whpalmer4

Yes, that's it, I think.

32. ParthKohli

And the solutions can't exceed $$15$$.

33. klimenkov

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

34. 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.

35. ParthKohli

What's that? Mathematica?

36. klimenkov

Yes.

37. ParthKohli

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

38. 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

39. whpalmer4

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

40. ParthKohli

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

41. ParthKohli

Work on this*

42. 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!

43. klimenkov

Ok. I'll try. Good night.

44. LOOSEr

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

45. LOOSEr

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

46. 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...

47. LOOSEr

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

48. LOOSEr

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

49. 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.