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ParthKohli
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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?
 one year ago
 one year ago
ParthKohli Group Title
How many four digit numbers are there such that the digits add up to 10 assuming that all digits are allowed to be zero?
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
less than 9999 but more than 1000 :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
1117 1171 1711 7111 1126 1162 1261 1621 2611 1216 1612 2161 6121 6211 2116 6112 etc
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
A more mathematical approach, probably?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
1000A + 100B + 10C + D = 10 100A + 10B + C + D/10 = 1
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
if the number is ABCD ^
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
pfft, that is the most basic mathematical appraoach i know :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :(
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
1000A + 100B + 10C + D = 10  (1) A + B/10 + C/100 + D/1000 = 1/100  (2)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
But we must have four equations, no?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
No problem :) Parth , thanks though..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I am going to subtract the above 2 eqns
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
its most likely a counting application with permutations since order matters
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Can we use combinatorics for this?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
hmn yes ^^ wait
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
sorry to correct what @mathslover is doing is a bit wrong /////// A+B+C+D=10 is the requirement///////////
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
1117; how many ways are there to permute this? 1126 how many ways? 1135 1144
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
we have to account for 0 i spose, which is added computations
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
OH yes @harsh314 thanks a lot
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
4+12+12+12 +things with zeros
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
118: stuff like 1180, 1108, 1018, but not 0118 127 136 145 19 28 37 46 55
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
so permute the 3 groups and times it by 3
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
1900, 1090, 1009 looks to be 3 times the 2 groups as well except for the 55 :) 5500, 5050, 5005 is just 3
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
This is very long. :
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Instead of bruteforce, can we try something simpler?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I got it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
this isnt brute force anymore, its refining
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Lemme check my method again
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Forget it . . . I think I must try it again :(
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
im to focused on my exam in 30 minutes to think thru this clearerer
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Parth, do you have answer of it ? ?
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
maybe closer the answer
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
84 numbers ... I think
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I will put up soln here soon... just thinking of more no.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
can the digits repeat Parth?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Well, a number still has four digits if the digits repeat. :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
5193 is the answer or am i wrong as usual
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Hey wait, did I mention that the first digit can be zero?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Oh! Is it allowed to have 0910 as the number ?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the first digit cant be zero and it still be a 4digit number tho
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
no but a number with first digit as 0 does not make sense look at my answer plz i think it is right
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
^ yeah after that 0910 is not a four digit number , it will be three digit number ..
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
have fun, i gotta run ;)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I know... the question is making such an assumption, so...
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 ;)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
well there can't be more than 186 numbers.. I think.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh sorry 186
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Is it 186?
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
more than 577
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
yes it is 186 @ParthKohli
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
The answer is 186?
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
(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 :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
yes @ParthKohli
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
186 when 0 is allowed or not?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
allowed
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
How? :S
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
282 and 186 are both not right :
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
i think i got the answer.............
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
282 was with leading zeros, without it would be 219
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
408 without the zeros
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
468 with zeroes
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
plz do reply if wrong.......
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I'd check
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Done! 13 C 3 = 286
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
yes i got this too a silly error how did you do it............
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
my method is tooo long.......
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
... 286 is the answer
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
i too got the same but by long method how did you get it .........
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
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 \(\binom{10 + 4  1}{4  1}\)
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
oh so no logic behind it as such..............
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I don't know. :
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
works in google chrome :) <html> <body> <input type=button value="AddUp" onClick="qwe()"><br> 0000: <input id="txtOut0"><br> 1000: <input id="txtOut1"> </body> <script language=javascript> var a=0,b=0,c=0,d=0 var count=0 function qwe() {for (var x=1000; x<=10000; x=x+1) {if (a+b+c+d == 10){count=count+1} d=d+1; if (d>10) {d=0; c=c+1} if (c>10){c=0; b=b+1} if (b>10){b=0; a=a+1} if (a >9){x=10001} } document.getElementById("txtOut0").value=count //resets a=1,b=0,c=0,d=0, count=0 for (var x=1000; x<=10000; x=x+1) {if (a+b+c+d == 10){count=count+1} d=d+1; if (d>10) {d=0; c=c+1} if (c>10){c=0; b=b+1} if (b>10){b=0; a=a+1} if (a >9){x=10001} } document.getElementById("txtOut1").value=count } </script> </html>
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
I'm working on one :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
if abcd = 10 then reset to 0, not if greater than 10 .... lol
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
im missing 9001, 9010, 9100 from the end of my 0xxx for some reason 279+3 = 282 otherwise its 219 if we use actual 4digit numbers :)
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
282 is correct :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
yay!! :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
... brute math, the easy way :)
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
Computers made these problem a lot easier :)
 one year ago

Meepi Group TitleBest ResponseYou've already chosen the best response.2
Unfortunately Parth has 286 as answer though :(
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
parth is counting somethings more than once then. i adjusted my loops to 0 to 10000 and it caught the last 3 ;)
 one year ago
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