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kondor19780726 [428]
3 years ago
9

An n-digit number is a positive number with exactly n digits. Nine hundred distinct n digit numbers are to be formed using only

three digit 5, 7, 9. The smallest value of n for which this is possible is :
(A) 6
(B)7
(C)8
(D)9
Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

\underbrace{3\cdot3\cdot3\cdot\ldots}_{n}\geq900\\\\
3^n\geq900\\\\
\log_3 3^n\geq \log_3900\\\\
n\geq \log_3900\\\\
n\geq\dfrac{\ln 900}{\ln 3}(\approx6.2)

So, the least n is 7.

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where <em>h</em> and <em>k</em> forms the coordinates of the centre of the circle.

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                        .                            x^2 \ + \ y^2 \ = r^2.

Now, we are interested in solving for the <em>x</em>-intercepts (the <em>x</em>-coordinates when the circle intersects the <em>x</em>-axis), of the circle

                                                      x^2 \ + \ y^2 \ = \ 4 .

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                                                     x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2.

Geometrically speaking, the tangent to the circle at the point defined by one of the <em>x</em>-intercepts of the circle is actually a vertical line, more specifically the lines x \ = \ \pm \ 2.

First and foremost, for the vertical line x \ = \ 2, it intersects the straight line y \ = \ \displaystyle\frac{1}{2}x  , giving the y-coordinate for point P,

                                                    y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1.

Hence, the coordinates of point P are (1, \ 1).

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Therefore, the coordinates of point P are also (1, \ -1).

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