Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Find all 3digit numbers such that the sum of the digits is 11 times less than the number.
 one year ago
 one year ago
Find all 3digit numbers such that the sum of the digits is 11 times less than the number.
 one year ago
 one year ago

This Question is Closed

AzteckBest ResponseYou've already chosen the best response.0
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?
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
I don't know how to correct that part. Is this a question from a worksheet or textbook?
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
This is my translation. You have to find all 3digit numbers, the sum of digits of which is 11 times less than the number. Can you write this correctly in English?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Find all 3digit numbers such that the sum of the digits is 11 times less than the number.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Actually, I used Google Translate, because I did not know the grammar.
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Ah okay, no worries mate.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Thank you @ParthKohli . Now solve this.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
So translating your question into an equation:\[\dfrac{100a + 10b + c}{11} = a + b + c\]Where the digits are \(a,b,c\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Another way to think is to see the multiples of \(11\), because we are sure that our threedigit number is a multiple of \(11\): 110  no. 121  no. 132  no. 143  no. 154  no. 165  no. Too long.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
You don't find easy ways...
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
0 is not a 3digit number @whpalmer4 .
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
\[11a + 11b + 11c = 100a + 10b + c \iff 89a + b + 10c = 0\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Yes. It is easier to find.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Yeah, let's try \(a = 1\). Then \(b = 9\) and \(c = 8\) which is exactly what @whpalmer4 said
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
If \(a = 2\), then \(b = 8\) but \(c > 9\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
So I guess \(198\) is the only number.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
(though it will toss out 000 as a solution)
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
I would give you both medals if i could.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Don't worry, whpalmer has mine.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
@whpalmer4 Don't worry with the 000. We can figure out trivial solutions in the end. :D
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
WolframAlpha writes the solution as \[b=9a + 10n, c = 8an, n \in Z\] which makes it clear that 198 is the only solution if a b and c all have to be single digits
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Fun little latenight problem :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Holy WolframAlpha!
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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? :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
\[17a + 17b + 17 c= 256a + 16b + c\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Yes, that's it, I think.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
And the solutions can't exceed \(15\).
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Do[If[11*(IntegerDigits[i][[1]] + IntegerDigits[i][[2]] + IntegerDigits[i][[3]]) == i, Print[i]], {i, 100, 999}] 198
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
What's that? Mathematica?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
So does that mean this: If we are looking at base \(n\), and the multiplier is \(n + 1\), then there's one solution.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
I'll check back in the morning to see what you guys have figured out :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.4
Nah, I am pretty bad at mathematics, I'd just leave this thread and we'd on this tomorrow. =)
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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!
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Ok. I'll try. Good night.
 one year ago

LOOSErBest ResponseYou've already chosen the best response.0
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
 one year ago

LOOSErBest ResponseYou've already chosen the best response.0
So, 9*10*10+1*10*8+1*9*1
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.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...
 one year ago

LOOSErBest ResponseYou've already chosen the best response.0
No.... what I mean is Number of 3digit numbers such that the sum of the digits is 11 times less than the number = 989
 one year ago

LOOSErBest ResponseYou've already chosen the best response.0
Sorry, it is 8*10*10+1*10*8+1*9*1=889
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
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.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.