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[Fwd: Selecting numbers with all odd digits]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg22274] [Fwd: [mg22170] Selecting numbers with all odd digits]
*From*: Hartmut Wolf <hwolf at debis.com>
*Date*: Mon, 21 Feb 2000 20:00:42 -0500 (EST)
*Organization*: debis Systemhaus
*Sender*: owner-wri-mathgroup at wolfram.com
Hartmut Wolf schrieb:
Tom De Vries schrieb:
>
> Hello all,
>
> In one of the math classes I am teaching we are doing some work with
> permutations and combinations. One question in the text was, "how many
> positive integers less than 700 have all odd digits?"
>
> I had done the question "by hand" and then wondered how to use Mathematica
> to demonstrate that the solution was the correct one.
>
> I am still very far down on the learning curve in most things, but I managed
> to bludgeon my way to an answer (thankfully the same as mine!)
>
> Select[
>
> Range[699],
>
> Union[ OddQ [ IntegerDigits[#]]] == {True} &
>
> ]
>
> I am sure my trial and error method can be improved upon. I am not talking
> speed as much as simply using Mathematica in a way that makes sense and
> would be more logical??
>
Hello Tom,
as for the programming, your
In[1]:= Select[Range[699], Union[OddQ[IntegerDigits[#]]] == {True} &] //
Length
Out[1]= 105
is equivalent to
In[2]:= Count[Range[699], _?(And @@ OddQ[IntegerDigits[#]] &)]
Out[2]= 105
So this counts all Integers in the specified range for that all digits
are odd -- literal Translation. But you also could calculate simply
In[3]:= 5^^300 + 5^^100 + 5^^10
Out[3]= 105
or if you prefer
In[4]:= 3 5 5 + 5 5 + 5
Out[4]= 105
The example is somewhat very special, perhaps tricking. Better try to
solve the general case "by hand". After the experience with that you
might end up with that algorithm:
caodd3[n_Integer?Positive] :=
Block[{aodd = True,
f = #1*5 + If[aodd, If[OddQ[#2], (#2 + 1)/2,
aodd = False; #2/2], 5] & },
Fold[f, 0, IntegerDigits[n]]]
In[9]:= caodd3[699]
Out[9]= 105
To give you a hint how I discovered this, I took the arbitrary number
13461:
(1) caodd[13461] == caodd[13399]
(2) 1_3_3_9_9
1 5 5 5 5 +
2 5 5 5 +
2 5 5 +
5 5 +
5
The new algorithm has quite a different efficiency. Compare with
caodd2[n_Integer?Positive] :=
Count[Range[n], _?(And @@ OddQ[IntegerDigits[#]] &)]
In[13]:= caodd2[315225] // Timing
Out[13]= {101.816 Second, 7305}
In[14]:= caodd3[315225] // Timing
Out[14]= {0.01 Second, 7305}
Kind regards, Hartmut
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