Latest Elitmus Aptitude Question SOLUTION: If a number x^100 have 31 digit then how many digits are there in x^1000?

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If a number x^100 have 31 digit then how many digits are there in x^1000?

Asked In Elitmus Rozzer (10 years ago)

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RAKESH  (9 years ago)

x^100 have 31 digits => x = 2 only as 3^100 have more than 31.
check, if
y = 2^100
=> logy = 100*log2
=> logy = 100*.301
=> logy = 30.1
=> y has 30+1= 31 digits.

so, x^1000 = 2^1000 = y
=> logy = 1000*log2
=> logy = 301
=> y has 301+1 = 302 digits.

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vamsi  (10 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

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ajit rathore  (9 years ago)

y=x^100 [where y is 31]
taking log both side
log y = 100 log x
log y = 100 log(0.3010) [assume the value of x]
log y = 30.1 = 31 digit [value of x = 2 ]
then
y = 2 ^ 1000
taking log both side
log y = 1000 log 2
log y = 1000*0.3010 = 301 (ans)

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Nalini Konapala  (9 years ago)

i think 310

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Amit kumar jha  (10 years ago)

I cant understand your question can u give me option ,,,,

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yogita pal  (10 years ago)

@VAMI : This ans wasn't thr in the option.

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Akshay Mathur  (10 years ago)

Please enter the options as well..!!

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