Question

Square root of abcdefghi, where a ... I are digits,
will contain x digits. The value of x is
(1) 3
(2) 4
(3) 5
(4) 6​

Answer 1

Answer:

(3) 5

Explanation:

We know that, the ceiling function of the logarithm of a number to the base 10 gives the no. of digits of that number.

→  The ceiling function of a real number is the least integer greater than or equal to that number.

→  E.g.: Ceiling function of 3.26 is 4.

→  Ceiling function of $\mathsf{x}$ is denoted by $\mathsf{\lceil x\rceil}.$

→  $\sf{\lceil x-1\rceil\ \ \textless\ \ x\ \leq\ \lceil x\rceil}$

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Here we have to find $\sf{x=\left\lceil\log\left[(\it{abcd efghi})^{\sf{\frac{1}{2}}}\right]\right\rceil}.$

The number 'abcdefghi' consists of 9 digits. Thus the logarithm of this number to the base 10 has a floor function 9. i.e.,

$\mathsf{\lceil\log(\it{abcde fghi})\rceil\ \sf{=9}}$

Then,

$\sf{8\ \textless\ \log(\it{abcde fghi})\ \sf{\leq9}}$

Taking half of each,

$\mathsf{\dfrac {8}{2}\ \ \textless\ \ \dfrac {1}{2}\log(\it{abcd efghi})\ \leq\ \sf{\dfrac {9}{2}}}\\\\\\\mathsf {4\ \textless\ \log\left [(\it{abcd efghi})^{\sf{\frac {1}{2}}}\right]\leq\ 4.5}$

This implies,

$\mathsf{\left\lceil\log\left [(\it{abcd efghi})^{\sf{\frac {1}{2}}}\right]\right\rceil=}\ \mathbf{5}$

That is,

$\Large\textsf{x = }\Large\mathbf{5}$

Thus the square root of 'abcdefghi' has 5 digits.

Hence (3) is the answer.