First one to get it right will get best answer.
The sum of the digits 2004 equal 6.
How many digits from 1000 to 9999 equal 6?
2006-07-24 19:36:24 · 14 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Eulercrosser is right with his/her answer.
It comes 56.
But interestingly, if u go one step further, and consider for example 1059=15=6 or 1599=24=6 or 6999=33=6, then they r in total EXACTLY 1000 (56+944).
Interesting answer, isn't it..!!!
i wud b crazy to list all of them here. But if u want to know how i did that and wat r those numbers, add me brucelee_lover@yahoo.co.uk...;), i will send it to u...:-)..
2006-07-24 20:20:21 · answer #1 · answered by Xtreem 2 · 0⤊ 0⤋
Actually, the way you asked it, the answer would be 0, since a digit is a number between 0 and 9. Also you said equal 6, also not possible. There are no such numbers between 1000 and 9999, so the answer is 0.
If you want to know how many integers between 1000 and 9999 have the sum of their digits equal to 6, then the answer would be 56 numbers do:
1005
1014
1023
1032
1041
1050
1104
1113
1122
1131 (10)
1140
1203
1212
1221
1230
1302
1311
1320
1401
1410 (20)
1500
2004
2013
2022
2031
2040
2103
2112
2121
2130 (30)
2202
2211
2220
2301
2310
2400
3003
3012
3021
3030 (40)
3102
3111
3120
3201
3210
3300
4002
4011
4020
4101 (50)
4110
4200
5001
5010
5100
6000 (56)
2006-07-25 02:50:13 · answer #2 · answered by Eulercrosser 4 · 0⤊ 0⤋
56 is correct... but I will calculate it using a general method :)
I will count in how many ways we can write 6 as an ordered sum of numbers 0..6. This is equal to the numbers of ways in which we can place three "separators" between 6 objects, e.g.
* | * * || * * * = 1 2 0 3
| * * * * | * * | = 0 4 2 0 etc.
The separators can be placed in 7 positions; the number of combinations of 3 out of 7 with possible repetition equals
9*8*7 / 3*2*1 = 84
There are 84 numbers with sum of digits 6 in the range 0000 ... 9999.
To exclude the numbers starting in zero, we forbid separators in the position in front of the six objects. Now there are six possible positions, and the solution becomes
8*7*6 / 3*2*1 = 56
2006-07-25 03:46:28 · answer #3 · answered by dutch_prof 4 · 0⤊ 0⤋
1000
2006-07-25 04:58:31 · answer #4 · answered by pabols 2 · 0⤊ 0⤋
if we want all numbers that the sum of their four places-or less equal six
the number of numbers
=(6-1+4)C(4-1)=(9*8*7)/(1*2*3)
=84
but we want only those who are greater than 1000
then we have to ignore all who have three places or less
the number=9c3-(6-1+3)c(3-1)=84-28
=56
2006-07-25 06:50:14 · answer #5 · answered by mohamed.kapci 3 · 0⤊ 0⤋
Strange question.. First of all, how does the first sentence help us?? )) You are not asking about sum, but about just digits.
And second, there are no digits between 1000 and 9999, there are numbers only, so please redefine your question...
2006-07-25 02:41:18 · answer #6 · answered by Synaps 2 · 0⤊ 0⤋
53
2006-07-25 03:13:43 · answer #7 · answered by The Q-mann 3 · 0⤊ 0⤋
the answer is 1000
2006-07-25 09:57:37 · answer #8 · answered by pls help me 1 · 0⤊ 0⤋
only one, 2004
2006-07-25 05:02:52 · answer #9 · answered by para 3 · 0⤊ 0⤋
1005=>4!=24
1104=>4!=24
1113=>4!=24
1203=>4!=24
2004=>4!=24
3003=>4!=24
2202=>4!=24
2112=>4!=24
6000=>1=1
totally 8*24=192=1=193
2006-07-25 07:23:53 · answer #10 · answered by chella 1 · 0⤊ 0⤋