Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem C. 1170. (May 2013)

C. 1170. Agnes was born on 25 March and her brother Peter was born on 9 February. They would like to construct a third-degree function f of integer coefficients, such that f(9)=2 and f(25)=3. Is there such a function?

(5 pont)

Deadline expired on June 10, 2013.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Legyen \(\displaystyle f(x)=ax^3+bx^2+cx+d\), ahol \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) és \(\displaystyle d\) egész számok. Tudjuk, hogy \(\displaystyle x_1=9\), \(\displaystyle x_2=25\). Ekkor

\(\displaystyle f(x_2)-f(x_1)=(ax_2^3+bx_2^2+cx_2+d)-(ax_1^3+bx_1^2+cx_1+d)=\)

\(\displaystyle =(x_2-x_1)[a(x_2^2+x_1x_2+x_1^2)+b(x_2+x_1)+c].\)

Láthatóan a második tényező egész szám. A feltételek alapján \(\displaystyle f(x_2)-f(x_1)=3-2=1\), aminek oszthatónak kellene lenni \(\displaystyle (x_2-x_1)\)-gyel. Mivel \(\displaystyle x_2-x_1=16\), ezért ez lehetetlen.

Ilyen függvény tehát nem létezik.


Statistics:

97 students sent a solution.
5 points:88 students.
4 points:3 students.
3 points:5 students.
1 point:1 student.

Problems in Mathematics of KöMaL, May 2013