Question

b)Can you order the digits 1, 2, 3, 4, 5 and 6 so that they make a
number with these characteristics:
a six digit number so formed is divisible by 6, and
when the final/last digit is removed it becomes a five-digit number
divisible by 5, and when the last digit is removed again it becomes a
four-digit number divisible by 4, and when this is repeated it becomes
a three-digit number divisible by 3, and When it is repeated again it
becomes a two-digit number divisible by 2. Of course when it is
repeated for a last time it will naturally be one-digit number divisible
by 1.​

Answer 1

1st condition:

In order to be divisible by 6 it must and should be divisible by both 2 and 3.

To be divisible by 2, the last digit of the number must be even.

To be divisible by 3, the sum of the digits of the number must be divisible by 3.

So assume that the last digit of our six digit number may be 2 or  4 or 6 so that it is divisible by both 2 and 3.

2nd condition:

In order to be divisible by 5, the last digit should be 5 or 0.

We have no 0's .

So the last digit is 5 in our five digit number.

Now assume that the last two digits of our required six digit number may be 52 or 54 or 56.

3rd condition:

In order to be divisible by 4 the last two digits of our 4 digit number should be divisible by 4.

32 is divisible by 4

36 is divisible by 4.

So, our four digit number may end with 32 or 36.

Now assume that the last four digits of our required six digit number may be 3254 or 3256 or 3654 or 3652

4th condition:

In order to be divisible by 3 the sum of the digits of a number must be divisible by 3.

Assume our three digit number as 123.

Sum of the digits = 6 which is divisible by 3

From these 4 conditions we can assume our number as 123654.

So, our required six digit number is 123654