Difference between revisions of "1975 AHSME Problems/Problem 29"
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==Solution(The Real Way - Binomial Thereom)== | ==Solution(The Real Way - Binomial Thereom)== | ||
− | Let's evaluate <math>(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6 | + | Let's evaluate <math>(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6</math>. We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since <math>(\sqrt{3}-\sqrt{2})^6) < 1</math>, the greatest integer greater than our original expression is <math>\boxed {\textbf{(C) } 970}</math> |
==See Also== | ==See Also== | ||
{{AHSME box|year=1975|num-b=28|num-a=30}} | {{AHSME box|year=1975|num-b=28|num-a=30}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:50, 21 June 2023
Contents
[hide]Problem
What is the smallest integer larger than ?
Solution(Very Stupid)
Then, find that is about . Finally, multiply and add to find that the smallest integer higher is
Solution(The Real Way - Binomial Thereom)
Let's evaluate . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since , the greatest integer greater than our original expression is
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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