On Mon, Sep 20, 2010 at 1:15 PM, Baljeet Kumar <[email protected]>wrote:

> If a and b are the numbers then
> dig = log10(a) + log10(b);
> if dig has some fractional part then number of digits is dig + 1 else dig.
>
>
found this correct onw


>
> On Mon, Sep 20, 2010 at 11:19 AM, sumant hegde <[email protected]>wrote:
>
>>
>> Adding to the partial solution, if x, y are first digits, and  x*y + x + y
>> < 10, the result will be  a+b -1 digits. "If not, u will need a complex
>> logic to solve"
>
>
if we take 30 *  33 as an example then it is (3*3 + 3+3 )> 10  which says
ans will be 4 digit
but ans is 990 which is 3 digit.


>> On Mon, Sep 20, 2010 at 10:50 AM, rahul patil <
>> [email protected]> wrote:
>>
>>> A partial solution is , if you multiply first digits of  two nos  and
>>> result is greater than 10 then surely result will be a+b digits
>>> If not, according to me, u will need a complex logic to solve.
>>>
>>>
>>> On Mon, Sep 20, 2010 at 10:41 AM, Srinivas <
>>> [email protected]> wrote:
>>>
>>>> how to find the no. of digits in the product of two numbers without
>>>> multiplying??
>>>> if a is the number of digits in A and
>>>> if b is the number of digits in B
>>>> the number of digits in A*B is either a+b or a+b-1 but how to find the
>>>> exact one?
>>>>
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>>>
>>>
>>> --
>>> Regards,
>>> Rahul Patil
>>>
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-- 
Regards,
Rahul Patil

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