 Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL? # Solutions for exercises "C" in November, 2001

In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.

C. 645.  Two players play the following game. They take turns in taking matches from a heap that initially contains 7 matches. In each step, a player can take one, two or three matches. The game continues until there are no matches left. The winner is the player holding an even number of matches at the end. Which player has a winning strategy, the one who starts or his opponent? How should he play in order to win?

Kvant

Solution. As 7 is an odd number, only one player can win. The opening player has a winning strategy. Suppose he opens by taking 3 matches. If his opponent takes 1 match in the next step, he will win by taking the remaining 3. If his opponent removes 2 matches, he can take 1 of the remaining 2. Finally, if the opponent takes 3 matches, he has the remaining 1, and he wins with 4 matches. (See the solution of problem B. 3493.)

C. 646.  In the sequence obtained by omitting the squares from the sequence of natural numbers, what is the 2001st term? Which term of the sequence is the number 2001?

Proposed by: P. Nádor, Pécs

Solution. As 442=1936<2001<2025=452, there are 44 square numbers up to 2001. Hence 2001 is the 2001-44=1957th member of the sequence. The first square number following 2001 is 2025=452, then 462=2116. Thus 2001 is followed by 23 non-square numbers in a row, the largest of which is 2024, in the 1957+23=1980th place. Then 2115-2025=90 non-square numbers follow: 2026, 2027, ...; the 21th of which, 2026+20=2046 is the 2001st member of the sequence. (Notice the almost symmetrical positions of 1957 and 2046 in the sequence; this is no accident.)

C. 647.  We have graphed the function f(x)=1/x in the coordinate plane. We want to alter the units on the coordinate axes so that the curve should represent the function g(x)=2/x. How should the new unit be set if it is to be the same on both axes?

Solution. If a segment of length e is chosen as the new unit on both axes, the new coordinates of the point (a,b) will be $\displaystyle \left({a\over e},{b\over e}\right)$. For the old coordinates of the graph of the function $\displaystyle f(x)={1\over x}$, $\displaystyle \left(x,{1\over x}\right)$. Thus with the new units, $\displaystyle \left({x\over e},{1/x\over e}\right)$. This corresponds to the graph of the function $\displaystyle g(x)={2\over x}$ if and only if $\displaystyle {1/x\over e}={2\over x/e}$ (for every x$\displaystyle ne$0). The necessary and sufficient condition for the identity to hold is $\displaystyle {1\over e}=2e$, that is, $\displaystyle e={\sqrt{2}\over2}$.

C. 648.  Evaluate 2log618.3log63.

Solution. 2log618.3log63=2log66+log63.3log63=21+log63.3log63=2.2log63.3log63=2.(2.3)log63=2.6log63=2.3=6.

C. 649.  The area of the base of a truncated pyramid is 8 cm2, and the area of the cover is 1 cm2. The pyramid is cut with a plane parallel to the base into two parts of equal volumes. Find the area of the intersection.

Proposed by: Á. Besenyei, Tatabánya

Solution. Let the areas of the bases of a truncated pyramid be A and B where A is larger. Complete the truncated pyramid to form a pyramid. If the height of the truncated pyramid is m and that of the whole pyramid is m+x, then $\displaystyle (m+x):x=\sqrt{A}:\sqrt{B}$ from the similarity of the appropriate pyramids, and hence $\displaystyle x=m{\sqrt{B}\over\sqrt{A}-\sqrt{B}}$. Thus the volume of the complete pyramid is $\displaystyle V={m+x\over3}A={m\over3}{A\sqrt{A}\over\sqrt{A}-\sqrt{B}}$, and that of the truncated pyramid is $\displaystyle V\left(1-\left({x\over m+x}\right)^3\right)=V\left(1-{B\sqrt{B}\over A\sqrt{A}}\right)$. The area of the larger base of the "lower" truncated pyramid obtained by cutting the truncated pyramid into two is also A. Denote the area of its smaller base by C. As this truncated pyramid can be completed to the same pyramid, of volume V, its volume is $\displaystyle V\left(1-{C\sqrt{C}\over A\sqrt{A}}\right)$ by the above reasoning. The condition of the problem is therefore

$\displaystyle 1-{C\sqrt{C}\over A\sqrt{A}}={1\over2}\left(1-{B\sqrt{B}\over A\sqrt{A}}\right),$

, and the solution of that is $\displaystyle C=(A\sqrt{A}+B\sqrt{B})^{2/3}=\left({16\sqrt{2}+1\over2}\right)^{2/3}\approx5.19$.