Question

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

Answer 1

less than 9999 but more than 1000 :)

Answer 2

1117
1171
1711
7111

1126 1162 1261 1621 2611
1216 1612 2161 6121 6211
2116 6112

etc

Answer 3

A more mathematical approach, probably?

Answer 4

1000A + 100B + 10C + D = 10

100A + 10B + C + D/10 = 1

Answer 5

if the number is ABCD ^

Answer 6

pfft, that is the most basic mathematical appraoach i know :)

Answer 7

I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :-(

Answer 8

1000A + 100B + 10C + D = 10 ---- (1)
A + B/10 + C/100 + D/1000 = 1/100 ---- (2)

Answer 9

But we must have four equations, no?

Answer 10

No problem :) Parth , thanks though..

Answer 11

I am going to subtract the above 2 eqns

Answer 12

its most likely a counting application with permutations since order matters

Answer 13

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

Answer 14

Can we use combinatorics for this?

Answer 15

hmn yes ^^ wait

Answer 16

combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable

Answer 17

sorry to correct what @mathslover is doing is a bit wrong ///////
A+B+C+D=10 is the requirement///////////

Answer 18

1117; how many ways are there to permute this?

1126 how many ways?

1135

1144

Answer 19

we have to account for 0 i spose, which is added computations

Answer 20

OH yes @harsh314 thanks a lot

Answer 21

4+12+12+12

+things with zeros

Answer 22

118: stuff like 1180, 1108, 1018, but not 0118


127
136
145

19
28
37
46
55

Answer 23

so permute the 3 groups and times it by 3

Answer 24

1900, 1090, 1009 looks to be 3 times the 2 groups as well

except for the 55 :) 5500, 5050, 5005 is just 3

Answer 25

This is very long. :-|

Answer 26

Instead of brute-force, can we try something simpler?

Answer 27

I got it

Answer 28

this isnt brute force anymore, its refining

Answer 29

Lemme check my method again

Answer 30

Forget it . . . I think I must try it again :(

Answer 31

im to focused on my exam in 30 minutes to think thru this clearerer

Answer 32

Parth, do you have answer of it ? ?

Answer 33

Nope.

Answer 34

maybe closer the answer

Answer 35

84 numbers ... I think

Answer 36

How?

Answer 37

I will put up soln here soon... just thinking of more no.

Answer 38

can the digits repeat Parth?

Answer 39

Yes.

Answer 40

Well, a number still has four digits if the digits repeat. :-)

Answer 41

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?

Answer 42

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.

Answer 43

5193 is the answer or am i wrong as usual

Answer 44

Hey wait, did I mention that the first digit can be zero?

Answer 45

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?

Answer 46

Oh! Is it allowed to have 0910 as the number ?

Answer 47

Yes.

Answer 48

the first digit cant be zero and it still be a 4-digit number tho

Answer 49

no but a number with first digit as 0 does not make sense look at my answer plz i think it is right

Answer 50

^ yeah after that 0910 is not a four digit number , it will be three digit number ..

Answer 51

have fun, i gotta run ;)

Answer 52

I know... the question is making such an assumption, so...

Answer 53

@amistre64 ;)

Answer 54

well there can't be more than 186 numbers.. I think.

Answer 55

^ 196

Answer 56

oh sorry 186

Answer 57

Is it 186?

Answer 58

more than 577

Answer 59

yes it is 186 @ParthKohli

Answer 60

The answer is 186?

Answer 61

(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 :)

Answer 62

yes @ParthKohli

Answer 63

186 when 0 is allowed or not?

Answer 64

allowed

Answer 65

How? :S

Answer 66

282 and 186 are both not right :-|

Answer 67

i think i got the answer.............

Answer 68

282 was with leading zeros, without it would be 219

Answer 69

408 without the zeros

Answer 70

468 with zeroes

Answer 71

plz do reply if wrong.......

Answer 72

@ParthKohli

Answer 73

I'd check

Answer 74

Done! 13 C 3 = 286

Answer 75

yes i got this too a silly error how did you do it............

Answer 76

my method is tooo long.......

Answer 77

@ParthKohli

Answer 78

... 286 is the answer

Answer 79

i too got the same but by long method how did you get it .........

Answer 80

I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4

Answer 81

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 \(\binom{10 + 4 - 1}{4 - 1}\)

Answer 82

oh so no logic behind it as such..............

Answer 83

But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?

Answer 84

I don't know. :-|

Answer 85

Weird :l

Answer 86

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

Answer 87

but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............

Answer 88

@satellite73

Answer 89

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>

Answer 90

i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected

Answer 91

I'm working on one :)

Answer 92

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

Answer 93

if abcd = 10 then reset to 0, not if greater than 10 .... lol

Answer 94

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 :)

Answer 95

282 is correct :)