Mathematical and Physical Journal
for High Schools
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Problem C. 1394. (January 2017)

C. 1394. How many positive integers are there whose prime factorization contains only the two smallest primes, and whose third power has eight times as many positive divisors as the number has?

(Matlap, Kolozsvár)

(5 pont)

Deadline expired on February 10, 2017.


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

Megoldás. A keresett számok \(\displaystyle 2^p\cdot3^q\) alakúak, ahol \(\displaystyle p,\,q\in \Bbb N^+\).

Az ilyen alakú számok pozitív osztóinak száma: \(\displaystyle (p+1)(q+1)\).

A számok harmadik hatványa \(\displaystyle 2^3p\cdot 3^3q\) alakú.

Ezek pozitív osztóinak száma: \(\displaystyle (3p+1)(3q+1)\).

A feltétel alapján \(\displaystyle 8(p+1)(q+1)=(3p+1)(3q+1)\).

Az egyenletet rendezve:

\(\displaystyle 8pq+8p+8q+8=9pq+3p+3q+1,\)

\(\displaystyle 0=pq-5p-5q-7=(p-5)(q-5)-32.\)

Tehát a \(\displaystyle (p-5)(q-5)=32\) egyenletet kaptuk.

32 lehetséges osztóit figyelembe véve a következő megoldások adódnak:

\(\displaystyle (p-5,q-5)\in\{(1,32),(2,16),(4,8),(8,4),(16,2),(32,1)\},\)

amiből

\(\displaystyle (p,q)\in\{(6,37),(7,21),(9,13),(13,9),(21,7),(37,6)\}.\)

Tehát 6 ilyen szám van.


Statistics:

194 students sent a solution.
5 points:149 students.
4 points:6 students.
3 points:7 students.
2 points:20 students.
1 point:7 students.
0 point:4 students.
Unfair, not evaluated:1 solution.

Problems in Mathematics of KöMaL, January 2017