A Peculiar Number!

BCD mod 5 = 0; therefore D = 5

ABC mod 4 = 0 --> BC mod 4 = 0 BC has two possible values: 12 and 24

If we take BC = 12, ABCDE will be A125E, with 3 and 4 unused. However, E cannot be 3 or 4 since C+D+E must be divisible by 3. So, BC = 24

ABCDE = A245E

E must be 3 so 4+5+3 = 12 which is divisible by 3, leaving A = 1

Answer: 12453

You have 5 spaces allotted: _ _ _ _ _

4th place must be 5, since it is divisible by 5.

You now have:

_ _ _ 5 _

2nd and 3rd place can be (a)12, (b)24, (c)32 or (d)52.

But the sum of the last three places must equal something divisible by 3

(a)2+5+2 is 9, but no repeating is allowed

(b)4+5+3 is 12, so this is a possibility

(c)2+5+2 is 9, again not possible

(d)You cannot use the 5 again, since it is definitely used in space 4

You now have _ 2 4 5 3

The remaining space is 1, since it is the only number not used. Therefore, the number is 1 2 4 5 3.

Since BCD is divisible by 5, D must be 5. Since ABC is divisible by 4, C must be 2 or 4. Since CDE is divisible by 3, C+D+E is a multiple of 3. If C is 2, then E would have to be 2 (making 9) or 5 (making 12). Both 2 and 5 have already been used, so there is a contradiction. So C cannot be 2. Thus, C is 4. E must be 3, as that is the only way C+D+E can be a multiple of 3. So AB must be 12 or 21. The only solution that works is one that makes ABC a multiple of 4, and that is 124. Thus, the answer is 12453.

ABC can be ?12 or ?24

BCD can be 1?5 or 2?5 -> D = 5

CDE can be only 453 -> C = 4, E = 3

ABC can be only 124 -> A = 1, B = 2 -> ABCDE = 12453

Start with ABC. It must be divisible by 4, so its last two digits must both be divisible by four. Keep in mind what the other 2 number combinations need to end in, since BCD has to divide by 5 evenly and CDE 3. You can come up with 124 for ABC- 2 and 4 are both divisible by 4, so it works.
BCD is divisible by 5, so its last digit, D, must be 5 or 0. Input the 5.
CDE must be divisible by 3, and the only number left is 3. You can check your work- the three divisibility rule is that if the digits add up to a multiple of 3, the number is divisible by 3. CDE would be 453, and 4+5+3=12. 12 is a multiple of 3, so 453 is divisible by 3.

Here was my method:

We have 5 spots: _____

The 2nd and 3rd ones combined can only be 4 numbers: 12 24 32 54

The 4th spot has to be 5 so now we've removed one number, leaving us: 12 24 32

2+5+2=9, which is divisible by 9 but we used 2 2's. We can only use 1 2. Now we've cut it down to just one: 24

4+5+x=12, meaning x=3, meaning it should be the last digit.

So now we have _2453. The only digit we have left is 1. Therefore, the number is: 12453

We have to use the numbers 1,2,3,4 and 5 to make a five digit number. The first requirement is that the first three digits form a number dvisible by four, which can only be achieved from these numbers by using 124 (= 31 x 4), then digits 2,3 and 4 must form a number divisible by five, so the fourth digit has to be 5 as numbers divisible by five end either in five ior zero and zero is not available to us. That leaves us the fifth digit to fill, and the only number we have not used is 3, hence the number is 12,453, and back checking using the last limitation, that the final three digits be divisible by three confirms this (453 = 151 x 3).

Simple, the ABC's condition has only one number - 24 in the end. So ABC = 124. Next, BCD's condition has only one end - 5, so the BCD = 245. Finally, the last number is 3. Therefore, ABCDE = 12453.