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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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?

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

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0This 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?

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

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, I used Google Translate, because I did not know the grammar.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah okay, no worries mate.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you @ParthKohli . Now solve this.

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4Another 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.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0You don't find easy ways...

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.00 is not a 3digit number @whpalmer4 .

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4\[11a + 11b + 11c = 100a + 10b + c \iff 89a + b + 10c = 0\]

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. It is easier to find.

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4If \(a = 2\), then \(b = 8\) but \(c > 9\).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4So I guess \(198\) is the only number.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1for (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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1(though it will toss out 000 as a solution)

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0I would give you both medals if i could.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4Don't worry, whpalmer has mine.

ParthKohli
 3 years ago
Best 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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1WolframAlpha 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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Fun little latenight problem :)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1So, here's my parting shot — what about a similar problem, except in base 16, and the multiple is 17 * the sum of the digits? :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4\[17a + 17b + 17 c= 256a + 16b + c\]

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, that's it, I think.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4And the solutions can't exceed \(15\).

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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Here'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.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.4What's that? Mathematica?

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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1For 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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1I'll check back in the morning to see what you guys have figured out :)

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

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1I'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!

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. I'll try. Good night.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.011(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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, 9*10*10+1*10*8+1*9*1

whpalmer4
 3 years ago
Best 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...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, it is 8*10*10+1*10*8+1*9*1=889

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Ah, 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.
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