Australian Mathematics Competition 2017 Intermediate Q30

Algebra Level 4

Let G = 1 0 100 , G=10^{100}, and let n n be the largest integer satisfying n n < 1 0 G . n^n < 10^G.

How many digits does n n have?


The answer is 99.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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2 solutions

Sharky Kesa
Jul 26, 2017

Firstly, 1 0 G = 1 0 1 0 100 = ( 1 0 100 ) 1 0 98 10^G = 10^{10^{100}} = (10^{100})^{10^{98}} . Thus, n > 1 0 98 n>10^{98} .

Furthermore, ( 1 0 99 ) 1 0 99 = 1 0 99 × 1 0 99 > 1 0 10 × 1 0 99 = 1 0 1 0 100 (10^{99})^{10^{99}} = 10^{99 \times 10^{99}} > 10^{10 \times 10^{99}} = 10^{10^{100}} .

Thus, n < 1 0 99 n<10^{99} , so 1 0 98 < n < 1 0 99 10^{98} < n < 10^{99} . Therefore, n n has 99 \boxed{99} digits.

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Did the same.

Aditya Kumar - 3 years, 9 months ago
Rohan Jasani
Aug 29, 2017

Taking log makes it very easy to find its range

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