Here's the question you clicked on:
klimenkov
Find all 3-digit numbers such that the sum of the digits is 11 times less than the number.
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?
I don't know how to correct that part. Is this a question from a worksheet or textbook?
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?
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.
So translating your question into an equation:\[\dfrac{100a + 10b + c}{11} = a + b + c\]Where the digits are \(a,b,c\).
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.
You don't find easy ways...
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.
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
(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
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
Fun little late-night problem :-)
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? :-)
\[17a + 17b + 17 c= 256a + 16b + c\]
Yes, that's it, I think.
And the solutions can't exceed \(15\).
Do[If[11*(IntegerDigits[i][[1]] + IntegerDigits[i][[2]] + IntegerDigits[i][[3]]) == i, Print[i]], {i, 100, 999}] 198
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.
What's that? Mathematica?
So does that mean this: If we are looking at base \(n\), and the multiplier is \(n + 1\), then there's one solution.
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
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. =)
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!
Ok. I'll try. Good night.
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
So, 9*10*10+1*10*8+1*9*1
@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...
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
Sorry, it is 8*10*10+1*10*8+1*9*1=889
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.