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How many four digit numbers are there such that the digits add up to 10 assuming that all digits are allowed to be zero?

Mathematics
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less than 9999 but more than 1000 :)
1117 1171 1711 7111 1126 1162 1261 1621 2611 1216 1612 2161 6121 6211 2116 6112 etc
A more mathematical approach, probably?

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Other answers:

1000A + 100B + 10C + D = 10 100A + 10B + C + D/10 = 1
if the number is ABCD ^
pfft, that is the most basic mathematical appraoach i know :)
I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :-(
1000A + 100B + 10C + D = 10 ---- (1) A + B/10 + C/100 + D/1000 = 1/100 ---- (2)
But we must have four equations, no?
No problem :) Parth , thanks though..
I am going to subtract the above 2 eqns
its most likely a counting application with permutations since order matters
how many basic ways are there to define 4 digits that add up to ten, and then how many ways can each group be sorted thru
Can we use combinatorics for this?
hmn yes ^^ wait
combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable
sorry to correct what @mathslover is doing is a bit wrong /////// A+B+C+D=10 is the requirement///////////
1117; how many ways are there to permute this? 1126 how many ways? 1135 1144
we have to account for 0 i spose, which is added computations
OH yes @harsh314 thanks a lot
4+12+12+12 +things with zeros
118: stuff like 1180, 1108, 1018, but not 0118 127 136 145 19 28 37 46 55
so permute the 3 groups and times it by 3
1900, 1090, 1009 looks to be 3 times the 2 groups as well except for the 55 :) 5500, 5050, 5005 is just 3
This is very long. :-|
Instead of brute-force, can we try something simpler?
I got it
this isnt brute force anymore, its refining
Lemme check my method again
Forget it . . . I think I must try it again :(
im to focused on my exam in 30 minutes to think thru this clearerer
Parth, do you have answer of it ? ?
Nope.
maybe closer the answer
84 numbers ... I think
How?
I will put up soln here soon... just thinking of more no.
can the digits repeat Parth?
Yes.
Well, a number still has four digits if the digits repeat. :-)
I found this on net : http://answers.yahoo.com/question/index?qid=20080330145752AA36Ac2 Now, I am thinking can we use this method to find the answerr?
We can think of the ones with zeroes like this: three digit numbers that add up to 10 * 4! two digit numbers that add up to 10 * 4!/2 there can't be 3 or 4 zeroes.
5193 is the answer or am i wrong as usual
Hey wait, did I mention that the first digit can be zero?
1117 is unique, 4 11xy has 12, and there are 3 ways to chose xy, and no zero groups 4(1)+3(12) 118 is unique, there are 3 ways, and 3 zero groups 1xy has 6 ways, and there are 4 ways to choose xy, and 3 zero groups 3(3)+3(6(4)) 55 is unique, theee is ony 1 way, and 3 zero groups xy 2 ways, there are 4 ways to choose xy, and 3 zero groups 3(1) + 3(2(4)) is there a flaw here?
Oh! Is it allowed to have 0910 as the number ?
Yes.
the first digit cant be zero and it still be a 4-digit number tho
no but a number with first digit as 0 does not make sense look at my answer plz i think it is right
^ yeah after that 0910 is not a four digit number , it will be three digit number ..
have fun, i gotta run ;)
I know... the question is making such an assumption, so...
well there can't be more than 186 numbers.. I think.
^ 196
oh sorry 186
Is it 186?
more than 577
yes it is 186 @ParthKohli
The answer is 186?
(0..9).each do |a| (0..9).each do |b| (0..9).each do |c| (0..9).each do |d| count = count + 1 if (a + b + c + d) == 10 end end end end puts count gave me 282 as answer, I don't know a mathematical approach though :)
186 when 0 is allowed or not?
allowed
How? :S
282 and 186 are both not right :-|
i think i got the answer.............
282 was with leading zeros, without it would be 219
408 without the zeros
468 with zeroes
plz do reply if wrong.......
I'd check
Done! 13 C 3 = 286
yes i got this too a silly error how did you do it............
my method is tooo long.......
... 286 is the answer
i too got the same but by long method how did you get it .........
I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4
We wanted the number of non negative numbers \(a_1,a_2,a_3,a_4\) such that \(a_1 + a_2 + a_3 + a_4 = 10\). A formula says that it would be \(\binom{10 + 4 - 1}{4 - 1}\)
oh so no logic behind it as such..............
But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?
I don't know. :-|
Weird :l
my method was long and a bit calculation mistake.............. case1:::'1" as the first digit ,,,,,, the following subcases arise (a) .....with 2 zeroes ........9 has to be the sum of the remaining three numbers so total possible combinations are 3. (b) .......with 1 zero we have 810,720,630,540......all these have 6 combinations each so total 24 (c) .......with no zeroes.....711(3combinations),621(6),531(6),522(3),432(6),441(3),333(1) so total of 28 combinations hence for the 1st case we have 45 combinations similarly we can have others tooo..........and then add all........we get 286
but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............
works in google chrome :)
0000:
1000:
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i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected
I'm working on one :)
gotta weed out a bug :) 0,0,0,10 0,0,1,9 0,0,2,8 0,0,3,7 0,0,4,6 0,0,5,5 0,0,6,4 0,0,7,3 0,0,8,2 0,0,9,1 0,0,10,0 0,1,0,9 0,1,1,8
if abcd = 10 then reset to 0, not if greater than 10 .... lol
im missing 9001, 9010, 9100 from the end of my 0xxx for some reason 279+3 = 282 otherwise its 219 if we use actual 4-digit numbers :)
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282 is correct :)
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yay!! :)
... brute math, the easy way :)
Computers made these problem a lot easier :)
Unfortunately Parth has 286 as answer though :(
parth is counting somethings more than once then. i adjusted my loops to 0 to 10000 and it caught the last 3 ;)

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