MathDB

5

Part of 2009 Mathcenter Contest

Problems(2)

2^n can begin with any sequence of digits

Source: Mathcenter Contest / Oly - Thai Forum 2009 R1 p5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

11/9/2022
For n∈Nn\in\mathbb{N}, prove that 2n2^n can begin with any sequence of digits.
Hint: log⁡2\log 2 is irrational number.
number theoryDigitspower of 2
min P = (5a^{2}-3ab+2)/a^{2}(b-a)

Source: Mathcenter Contest / Oly - Thai Forum 2009 R2 p5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

11/9/2022
Let aa and bb be real numbers, where a≠0a \not= 0 and a≠ba \not= b and all the roots of the equation ax3−x2+bx−1=0ax^{3}-x^{2}+bx-1 = 0 is a real and positive number. Find the smallest possible value of P=5a2−3ab+2a2(b−a)P = \dfrac{5a^{2}-3ab+2}{a^{2}(b-a)}.
(Heir of Ramanujan)
inequalitiesalgebra