Latest Elitmus Aptitude Question SOLUTION: if no of digit in X^100 is 31.then find the number of digit in X^1000.

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(#M40018269) ELITMUS Numerical Ability question quant Keep an EYE

if no of digit in X^100 is 31.then find the number of digit in
X^1000.

Asked In Elitmus Deepak kumar Dubey (11 years ago)

Unsolved Read Solution (7)

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aruna  (11 years ago)

ans:301
if we have y digits in x^100, that means that
10^(y-1) = x^100 [we put 10 to the (y-1) power because 10^1 has]
[2 digits, 10^2 has 3 digits, so 10^(y-1) has ]
[y digits ]
in question we have 31 digits
10^(31-1)=x^100
applying log
30=100logx ------>equ 1

10^(y-1)=x^1000
y-1=1000 logx
y-1=10*100logx
y-1=10*30
y=301

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IFFAT IRAM  (11 years ago)

X^100 has 31 digits
no. of digits in a no. can be found by taking its log
i.e., log(X^100) = 31
=> 100 logX=31
multiplying both sides by ten
1000logX=310
=>
log(X^1000)=310
no. of digits = 310

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Vineet Pal  (11 years ago)

Answer is 302 digits.

10^(d−1) has d digits.
So for any number n, the number of digits of n is given by solving 10^(d−1)=n, or

d=1+⌊log10(n)⌋

put x=2. It is satisfying the given condition, that is,
1+100*⌊log10(2)⌋ = 1+100*⌊0.301⌋ = 1+⌊30.1⌋ = 1 + 30 = 31.

So
d = 1+1000*⌊log10(2)⌋ = = 1+1000*⌊0.301⌋ = 1+⌊301.something⌋ = 1 + 301 = 302

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babu rao ganpat rao apte  (10 years ago)

solution by aruna is correct... 10^(y-1)=x^100 is a formulae...where y will stand for number of digits

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praveen kumar  (11 years ago)

x^100=31 i.e (x^10)^10=31
x^1000=(((x^10)^10)^10)=(31^10)

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Ved Prakash Yadav  (9 years ago)

x^100 have 31 digits => x = 2 only as 3^100 have more than 31. but for x=2 no of digit will be 30
so, x^1000 = 2^1000 = y
=> logy = 1000*log2=1000*0.301
=> logy = 301
so y has 301+1 = 302 digits.

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Tejasvi Kumar  (6 years ago)

101 has 2 digits = 10
102 has 3 digits = 100
103 has 4 digits = 1000
1030 has 31 digits

therefore, 1030=x100
30=100logx
multiplying both sides by 10
30⋅10=10⋅100logx
300=1000logx=300+1=301

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