Can you help with this math puzzler?!!!?

Question

Find one of the 3 #'s being described by all of these clues.
4 digit # with no repeats
divisible by 5
the first 3 digits make a # divisible by 4
sum of all digits is 11
3 is in hundreds place.

Answer 1

2360

/ by five means that it ends in either 5 or 0.

3 in the hundreds, so look for a number divisable by four that is in the 30s

200, 400, and 800 are divisable by 4 and you can plug the 30 into them.

Answer 2

1325, 6320, 2360

Explanation:
you are given _ 3 _ _
_ 3 _ is divisible by 4, meaning 3 _ is divisible by 4 (a rule that is always true, for explanation see http://www.mathgoodies.com/lessons/vol3/divisibility.html)

so the original number is 1 of 2 possibilities:

1. _ 3 2 _
2. _ 3 6 _

since the number is divisible by 5, the last digit is either a 5 or a 0
(also see http://www.mathgoodies.com/lessons/vol3/divisibility.html for an explanation)

so the possibilities are

1. _ 3 2 5
2. _ 3 2 0
3. _ 3 6 0
( _ 3 6 5 cannot work, b/c sum of digits would be >11 )

since the sum of digits = 11,
just subtract the digits you know from 11 to get the thousands place, making the answers:
1. 1325,
2. 6320,
3. and 2360.

hope that helped! =)

Answer 3

Okay, this number has to end in a five or zero to be divisible by 5.

_3__ (3 is in the hundreds place, and no other number can be 3)

No number can be a 9, because that plus 3 would be 12, one more than 11.

No number can be 8, because that plus 3 would be 11, equaling 11 when you've got two dissimilar numbers to go.

The last number cannot be 5 because your thousands and tens numbers then have to equal three, and your only two combinations (3 is invalid, so you can't use 0) would have the first three digits add up to 6, non-divisible evenly by 4.

Your thousands number can therefore be 7, 6, 5, 4, 2, or 1.
Your hundreds number can only be 3.
Your tens number can therefore be 7, 6, 5, 4, 2, or 1.
Your ones number can only be 0.

_3_0

So of all possible combinations, start lower.

7310? (7+3+1+0=11; 731/4=182.75 -- not evenly)

6320 (6+3+2+0=11; 632/4=158; no repeats; 6320/5=1264)

6320 fits all of your criteria.

Answer 4

2360
1325
6320

*The ones digit must be a 5 or 0.
*If the first three digits are divisible by 4, then the tens digit must be a 2 or 6.
*There are then four possible last three digit combinations. 365 adds up to 14, so it is not possible. That leaves only three possible combinations. I subtracted the digits from 11 to come up with my thousands digit and, in turn, my answer.

Answer 5

1325

Answer 6

$a million.seventy 5. the 1st guy might might desire to put in $a million.seventy 5. Then there'd be $3.50 in the container, and after he's removing $2.00, there'd be $a million.50. Then the 2nd guy might put in $a million.50, so there'd be $3.00. After he takes his $2.00, there'd be $a million.00. The third guy might put in a greenback, making $2.00, then after he takes his $2.00, the container may well be emptied.

Answer 7

1325 is one of them.

Answer 8

2360

Answer 9

5320