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

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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1less than 9999 but more than 1000 :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.11117 1171 1711 7111 1126 1162 1261 1621 2611 1216 1612 2161 6121 6211 2116 6112 etc

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0A more mathematical approach, probably?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.01000A + 100B + 10C + D = 10 100A + 10B + C + D/10 = 1

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0if the number is ABCD ^

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1pfft, that is the most basic mathematical appraoach i know :)

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

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.01000A + 100B + 10C + D = 10  (1) A + B/10 + C/100 + D/1000 = 1/100  (2)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0But we must have four equations, no?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0No problem :) Parth , thanks though..

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0I am going to subtract the above 2 eqns

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1its most likely a counting application with permutations since order matters

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1how 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
 2 years ago
Best ResponseYou've already chosen the best response.0Can we use combinatorics for this?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.11117; how many ways are there to permute this? 1126 how many ways? 1135 1144

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1we have to account for 0 i spose, which is added computations

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0OH yes @harsh314 thanks a lot

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.14+12+12+12 +things with zeros

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1118: stuff like 1180, 1108, 1018, but not 0118 127 136 145 19 28 37 46 55

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1so permute the 3 groups and times it by 3

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.11900, 1090, 1009 looks to be 3 times the 2 groups as well except for the 55 :) 5500, 5050, 5005 is just 3

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0This is very long. :

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Instead of bruteforce, can we try something simpler?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1this isnt brute force anymore, its refining

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0Lemme check my method again

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0Forget it . . . I think I must try it again :(

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1im to focused on my exam in 30 minutes to think thru this clearerer

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0Parth, do you have answer of it ? ?

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0maybe closer the answer

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.084 numbers ... I think

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0I will put up soln here soon... just thinking of more no.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0can the digits repeat Parth?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Well, a number still has four digits if the digits repeat. :)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.0We 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.

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.05193 is the answer or am i wrong as usual

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Hey wait, did I mention that the first digit can be zero?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.11117 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
 2 years ago
Best ResponseYou've already chosen the best response.0Oh! Is it allowed to have 0910 as the number ?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1the first digit cant be zero and it still be a 4digit number tho

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1have fun, i gotta run ;)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I know... the question is making such an assumption, so...

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0well there can't be more than 186 numbers.. I think.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0yes it is 186 @ParthKohli

Meepi
 2 years ago
Best 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 :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0186 when 0 is allowed or not?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0282 and 186 are both not right :

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0i think i got the answer.............

Meepi
 2 years ago
Best ResponseYou've already chosen the best response.2282 was with leading zeros, without it would be 219

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0plz do reply if wrong.......

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0yes i got this too a silly error how did you do it............

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0my method is tooo long.......

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0... 286 is the answer

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0i too got the same but by long method how did you get it .........

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0We 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 \(\binom{10 + 4  1}{4  1}\)

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0oh so no logic behind it as such..............

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

harsh314
 2 years ago
Best ResponseYou've already chosen the best response.0my 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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1works 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>

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1gotta 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
 2 years ago
Best ResponseYou've already chosen the best response.1if abcd = 10 then reset to 0, not if greater than 10 .... lol

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1im 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 :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1... brute math, the easy way :)

Meepi
 2 years ago
Best ResponseYou've already chosen the best response.2Computers made these problem a lot easier :)

Meepi
 2 years ago
Best ResponseYou've already chosen the best response.2Unfortunately Parth has 286 as answer though :(

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1parth 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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