IIT JEE , AIEEE Forum - Mathematics , Algebra

prathima

A is the no. of digits in 4444 to the power of 4444

B is the no. of digits in A

then the no. of digits in B is ..................................

master_purav

This one looks simple..

Let x = (4444)^4444

So log x
= 4444 log 4444
= 4444 (3.6478)
= 16210.8

So number of digits in x = A= 16211

Now B is the number of digits in 16211 which is 5

Henve the answer is 5....

KraniuM

Dude what about the anti log?????

KraniuM

it should be like,

logx=16211

so x= antilog 16211
=is it 417000

so B=6

master_purav

@ KraniuM

There is no point in taking the antilog..

log x = 16210.8
which means that we take the antilog of 0.8 and palce the decimal after 1st number and multiply it by 10 to power 16210 which means whatever may be the answer it has 16211 digits.

Also B is the number of digits in 16211 ans not in its antilog.

So the answer should be 5...

KraniuM

Right on gotcha!!

Avinash_Bhat

Hey guys,

Truly, the no. of digits in B is 5. But the answer to the question is 1; since the qn. is to find out the no. of digits in B !!

avinash.sharma

Lets see the complete solution:

X= (4444) ⁴⁴⁴⁴

Log X = 4444 (log 4444)

= 4444 [log (4.444 x 10³)]

= 4444 [ 3log 10 + log 4.444]

= 4444 [3 + log 4.444]

= 3 x 4444 + 4444 x log 4.4444

= 13332 + 4444 x 0.8232

= 13332 + 3658.3008 = 16990.3008

Taking antilog

X = 4.782 x 10¹⁶⁹⁹⁰ = 0.4782 x 10¹⁶⁹⁹¹

A = number of digits in x = 16991

B = number of digits in A = 5

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A = number of digits in x = 16991

Then number of digits in B is only 1