Latest Elitmus Aptitude Question SOLUTION: If x^100 has 31 digit,then x^1000 will have how many digits?

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(#M40018614) ELITMUS Logical Reasoning question number Keep an EYE

If x^100 has 31 digit,then x^1000 will have how many digits?

Asked In Elitmus shash (11 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

shailesh  (11 years ago)

10^1 has 2 digits = 10
10^2 has 3 digits = 100
10^3 has 4 digits = 1000
similarly 10^y-1 has y digits
if we have y digits in x^100 that means
10^(y-1)=x^100
10^(31-1)=x^100
taking log both side
30 log 10 = 100 log x
30 = 100 log x
now,
10^(y-1)= x^1000
y-1=1000 log x
y-1=10*100 log x
put value of 100 log x in this equation
we get y-1=10*30
y=301
Ans=301

Is this solution Helpfull? Yes (96) | No (3)  

Sandeep S Hegde  (11 years ago)

31-1=100 log x solve this, we wil get x=2
so
2^1000 has (1000 log 2)+1=302
ANS: 302

Is this solution Helpfull? Yes (13) | No (6)  

pavan  (11 years ago)

302 digits

Is this solution Helpfull? Yes (7) | No (9)  

Umesh Yadav  (11 years ago)

10^1=(1+1)=2 digit
10^2=(2+1)=3 digit
.
.
.
10^30=(30+1)=31 digits
therefore, 10^30=x^100
30=100 log x
multiplaying both side by 10
30*10=10*100 log x
300= 1000 log x = 300+1=301

Is this solution Helpfull? Yes (4) | No (1)  

Sneha Agarwal  (11 years ago)

@pavan pls explain

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Deepak Chaudhari  (9 years ago)

The same question is repeated on 21st jun delhi center....and ans is 301.

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tech  (10 years ago)

i think 8 is wrong .......
answer is 12 ...
(team) (matches played) (win) (lose)
1 14 11 3
2 14 11 3
3 14 11 3
4 14 11 3
5 14 11 3
6 14 1 13
7 14 0 14
8 14 0 14

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Tarun Sharma  (9 years ago)

taking log so 100log(x)=30.abcd...(a belong to {0-9})
so log(x)=.30abcd...
so 1000 log(x)=30a.bcd
so no of digits [30a.bcd] +1 i.e 30a+1 so it can be 301 to 310

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Prateek Sancheti  (11 years ago)

@ Pavan can you please explain the answer?

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SAMBIT KUMAR GHOSH  (11 years ago)

please explain the soln...

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mohini gupta  (9 years ago)

302
solve using log

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