Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Algebra
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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.
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.
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 :-)
Holy Wolfram|Alpha!
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? :-)
\[17a + 17b + 17 c= 256a + 16b + c\]
That?
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?
Yes.
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. =)
Work on this*
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question