The Clock-Tower School Juniors Competition 3rd problem
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The Clock-Tower School Juniors Competition 3rd problem
Determinati toate perechile de numere naturale nenule \( a \) si \( b \) pentru care \( a^6\geq 5^{b+1} \) si \( b^6\geq 5^{a+1} \).
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Marius Mainea
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Pentru \( n\ge 6 \) natural,\( 5^{n+1}>n^6 \) si de aici a sau b este \( \le 5 \)
Last edited by Marius Mainea on Tue Mar 23, 2010 11:48 pm, edited 3 times in total.
De fapt, pt. orice \( n\in \mathbb{N} \) e adevarata inegalitatea aceea.
Observam ca daca \( a \le 5 \), atunci si \( b \le 5 \). Deci \( a, b \le 5 \) sau \( a, b \ge 6 \).
Daca \( a, b \le 5 \), obtinem \( (a,b)\in {(3;3),(4;4),(5;5)} \).
Daca \( a, b \ge 6 \), pp. ca \( a\le b \). Atunci \( a^6\ge 5^{b+1}>6^b\ge 6^a \) si rezulta ca \( a^6>6^a \), fals. (Deoarece daca \( x\ge y>e \), atunci \( x^y\le y^x \))
Observam ca daca \( a \le 5 \), atunci si \( b \le 5 \). Deci \( a, b \le 5 \) sau \( a, b \ge 6 \).
Daca \( a, b \le 5 \), obtinem \( (a,b)\in {(3;3),(4;4),(5;5)} \).
Daca \( a, b \ge 6 \), pp. ca \( a\le b \). Atunci \( a^6\ge 5^{b+1}>6^b\ge 6^a \) si rezulta ca \( a^6>6^a \), fals. (Deoarece daca \( x\ge y>e \), atunci \( x^y\le y^x \))