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.
On Mon, Sep 20, 2010 at 11:19 AM, sumant hegde sumant@gmail.com wrote:
Adding to the partial solution, if x, y are first digits, and x*y + x + y
10, the
On Mon, Sep 20, 2010 at 1:15 PM, Baljeet Kumar baljeetk...@gmail.comwrote:
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
@Naveen:
Yes you are right, the answer is not fully correct.
ans = integral part of ((log10(a)+log10(b) +1) {floor not ceiling}.
On Mon, Sep 20, 2010 at 4:11 PM, Naveen Agrawal nav.coo...@gmail.comwrote:
@Baljeet
I think your Answer is not fully correct
It should be :
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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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 lavudyasrinivas0...@gmail.comwrote:
how to find the no.
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
On Mon, Sep 20, 2010 at 10:50 AM, rahul patil rahul.deshmukhpa...@gmail.com
wrote:
A partial solution is , if you multiply first digits
