Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 AHSME Problems/Problem 29"

(→‎Solution(The Real Way - Binomial Thereom))
m (→‎Solution(The Real Way - Binomial Thereom))
 
Line 11: Line 11:
 
==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)</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>
+
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

Problem

What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$?

$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$

Solution(Very Stupid)

$(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$. Finally, multiply and add to find that the smallest integer higher is $\boxed {\textbf{(C) } 970}$

Solution(The Real Way - Binomial Thereom)

Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6$. 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 $(\sqrt{3}-\sqrt{2})^6) < 1$, the greatest integer greater than our original expression is $\boxed {\textbf{(C) } 970}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_AHSME_Problems/Problem_29&oldid=194666"