Polydivisble numbers using base b

Question

Picked the definition from wikipedia: A polydivisible number is a number with digits abcde... that has the following properties :

1. Its first digit a is not 0.

2. The number formed by its first two digits ab is a multiple of 2.

3. The number formed by its first three digits abc is a multiple of 3.

4. The number formed by its first four digits abcd is a multiple of 4.

   etc.

Will it continue, I mean. The number formed by its first five digits abcde is a multiple of 5, The number formed by its first six digits abcdef is a multiple of 6.... Then if it continues, is there any end point to it.

And

X number is a polydivisible in base y

For Instance

1232,

1 /1 = 1
12 /2 = 6
123 /3 = 41
1232 /4 = 308 

So, 1232 is a polydivisible number in base 4 and above

123220,

 1      /1 = 1             
 12     /2 = 6             
 123    /3 = 41            
 1232   /4 = 308           
 12322  /5 = 2464.4        
 123220 /6 = 220536.333r

So, 123,220 is not a polydivisible base 10 number,

Again 123,220,

 1      = 1       -> 1     /1 = 1      
 12     = 8       -> 8     /2 = 4      
 123    = 51      -> 51    /3 = 17     
 1232   = 308     -> 308   /4 = 77     
 12322  = 1850    -> 1850  /5 = 370    
 123220 = 11100   -> 11100 /6 = 1850  

 So, 123,220 is a `polydivisible` base 6 number (and a `polydivisible` base 10 number when converted to 11100 in base 10).

What do you mean X is a polydivisible number in base y. And also please help me to understand the above examples too.

A/c to first example what do you mean abovein polydivisible number in base y and above.

Is it possible to calculate nth polydivisible number using base b ?

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NOTE ####: I dont know which tag this question belongs too. Please pardon me if its a wrong tag and please bind the right tag to this question if you know it. Thanks

Answer 1

First, yes it can go on indefinitely.

Secondly, to clear up some confusion: In maths "base" refers to the number of values that a digit can take, e.g. base 10 is what we are used to (digits 0 to 9) and base 2 is binary.

1232 in base 4 gives:

1/1=1

12/2=3 # $12_4 = 6_{10}$

123/3=21 # $123_4 = 27_{10}$, $9_{10} = 21_4$

1232/4, well we can see that this one doesn't divide as all base 4 numbers have to end in 0 to be divisible by 4.

As such 1232 is not a polydivisible base 4 number.