ParthKohli
  • ParthKohli
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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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
less than 9999 but more than 1000 :)
amistre64
  • amistre64
1117 1171 1711 7111 1126 1162 1261 1621 2611 1216 1612 2161 6121 6211 2116 6112 etc
ParthKohli
  • ParthKohli
A more mathematical approach, probably?

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More answers

mathslover
  • mathslover
1000A + 100B + 10C + D = 10 100A + 10B + C + D/10 = 1
mathslover
  • mathslover
if the number is ABCD ^
amistre64
  • amistre64
pfft, that is the most basic mathematical appraoach i know :)
ParthKohli
  • ParthKohli
I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :-(
mathslover
  • mathslover
1000A + 100B + 10C + D = 10 ---- (1) A + B/10 + C/100 + D/1000 = 1/100 ---- (2)
ParthKohli
  • ParthKohli
But we must have four equations, no?
mathslover
  • mathslover
No problem :) Parth , thanks though..
mathslover
  • mathslover
I am going to subtract the above 2 eqns
amistre64
  • amistre64
its most likely a counting application with permutations since order matters
amistre64
  • amistre64
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
ParthKohli
  • ParthKohli
Can we use combinatorics for this?
mathslover
  • mathslover
hmn yes ^^ wait
amistre64
  • amistre64
combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable
anonymous
  • anonymous
sorry to correct what @mathslover is doing is a bit wrong /////// A+B+C+D=10 is the requirement///////////
amistre64
  • amistre64
1117; how many ways are there to permute this? 1126 how many ways? 1135 1144
amistre64
  • amistre64
we have to account for 0 i spose, which is added computations
mathslover
  • mathslover
OH yes @harsh314 thanks a lot
amistre64
  • amistre64
4+12+12+12 +things with zeros
amistre64
  • amistre64
118: stuff like 1180, 1108, 1018, but not 0118 127 136 145 19 28 37 46 55
amistre64
  • amistre64
so permute the 3 groups and times it by 3
amistre64
  • amistre64
1900, 1090, 1009 looks to be 3 times the 2 groups as well except for the 55 :) 5500, 5050, 5005 is just 3
ParthKohli
  • ParthKohli
This is very long. :-|
ParthKohli
  • ParthKohli
Instead of brute-force, can we try something simpler?
mathslover
  • mathslover
I got it
amistre64
  • amistre64
this isnt brute force anymore, its refining
mathslover
  • mathslover
Lemme check my method again
mathslover
  • mathslover
Forget it . . . I think I must try it again :(
amistre64
  • amistre64
im to focused on my exam in 30 minutes to think thru this clearerer
mathslover
  • mathslover
Parth, do you have answer of it ? ?
ParthKohli
  • ParthKohli
Nope.
anonymous
  • anonymous
maybe closer the answer
mathslover
  • mathslover
84 numbers ... I think
ParthKohli
  • ParthKohli
How?
mathslover
  • mathslover
I will put up soln here soon... just thinking of more no.
mathslover
  • mathslover
can the digits repeat Parth?
ParthKohli
  • ParthKohli
Yes.
ParthKohli
  • ParthKohli
Well, a number still has four digits if the digits repeat. :-)
mathslover
  • mathslover
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?
ParthKohli
  • ParthKohli
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.
anonymous
  • anonymous
5193 is the answer or am i wrong as usual
ParthKohli
  • ParthKohli
Hey wait, did I mention that the first digit can be zero?
amistre64
  • amistre64
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?
mathslover
  • mathslover
Oh! Is it allowed to have 0910 as the number ?
ParthKohli
  • ParthKohli
Yes.
amistre64
  • amistre64
the first digit cant be zero and it still be a 4-digit number tho
anonymous
  • anonymous
no but a number with first digit as 0 does not make sense look at my answer plz i think it is right
mathslover
  • mathslover
^ yeah after that 0910 is not a four digit number , it will be three digit number ..
amistre64
  • amistre64
have fun, i gotta run ;)
ParthKohli
  • ParthKohli
I know... the question is making such an assumption, so...
mathslover
  • mathslover
mathslover
  • mathslover
well there can't be more than 186 numbers.. I think.
mathslover
  • mathslover
^ 196
mathslover
  • mathslover
oh sorry 186
ParthKohli
  • ParthKohli
Is it 186?
anonymous
  • anonymous
more than 577
mathslover
  • mathslover
yes it is 186 @ParthKohli
ParthKohli
  • ParthKohli
The answer is 186?
anonymous
  • anonymous
(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 :)
mathslover
  • mathslover
yes @ParthKohli
ParthKohli
  • ParthKohli
186 when 0 is allowed or not?
mathslover
  • mathslover
allowed
ParthKohli
  • ParthKohli
How? :S
ParthKohli
  • ParthKohli
282 and 186 are both not right :-|
anonymous
  • anonymous
i think i got the answer.............
anonymous
  • anonymous
282 was with leading zeros, without it would be 219
anonymous
  • anonymous
408 without the zeros
anonymous
  • anonymous
468 with zeroes
anonymous
  • anonymous
plz do reply if wrong.......
anonymous
  • anonymous
@ParthKohli
ParthKohli
  • ParthKohli
I'd check
ParthKohli
  • ParthKohli
Done! 13 C 3 = 286
anonymous
  • anonymous
yes i got this too a silly error how did you do it............
anonymous
  • anonymous
my method is tooo long.......
anonymous
  • anonymous
@ParthKohli
ParthKohli
  • ParthKohli
... 286 is the answer
anonymous
  • anonymous
i too got the same but by long method how did you get it .........
anonymous
  • anonymous
I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4
ParthKohli
  • ParthKohli
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 \(\binom{10 + 4 - 1}{4 - 1}\)
anonymous
  • anonymous
oh so no logic behind it as such..............
anonymous
  • anonymous
But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?
ParthKohli
  • ParthKohli
I don't know. :-|
anonymous
  • anonymous
Weird :l
anonymous
  • anonymous
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
anonymous
  • anonymous
but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............
anonymous
  • anonymous
@satellite73
amistre64
  • amistre64
works in google chrome :)
0000:
1000:
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amistre64
  • amistre64
i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected
anonymous
  • anonymous
I'm working on one :)
amistre64
  • amistre64
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
amistre64
  • amistre64
if abcd = 10 then reset to 0, not if greater than 10 .... lol
amistre64
  • amistre64
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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anonymous
  • anonymous
282 is correct :)
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amistre64
  • amistre64
yay!! :)
amistre64
  • amistre64
... brute math, the easy way :)
anonymous
  • anonymous
Computers made these problem a lot easier :)
anonymous
  • anonymous
Unfortunately Parth has 286 as answer though :(
amistre64
  • amistre64
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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