1975 AMC 12 Problem 29

Attempt Problem 29 of the 1975 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1975 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

29.

What is the smallest integer larger than (3+2)6?(\sqrt3+\sqrt2)^6?
972972
971971
970970
969969
968968
Answer: C
Concepts:radicalbinomial theoremfloor and ceiling functions
Small hint:
Pair the expression with its conjugate (3−2)6(\sqrt3-\sqrt2)^6
Big hint:
Their sum is an integer, while the conjugate term lies strictly between 00 and 11
Solution:
Let α=3+2\alpha=\sqrt3+\sqrt2 and β=3−2.\beta=\sqrt3-\sqrt2. Adding their sixth powers cancels all odd radical terms: α6+β6=2(33+15(32)(2)+15(3)(22)+23)=970. \begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970. \end{aligned} Since 0<β<1,0\lt\beta\lt1, we have 969<α6=970−β6<970.969\lt\alpha^6=970-\beta^6\lt970. Thus the smallest larger integer is 970.970.
Therefore, the correct answer is C.
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