Problem B. 5295. (February 2023)

B. 5295. Find the largest integer \(\displaystyle k\) for which \(\displaystyle 1722\) divided by \(\displaystyle k\) leaves a remainder of \(\displaystyle 2m\) and \(\displaystyle 2179\) divided by \(\displaystyle k\) leaves a remainder of \(\displaystyle 3m\) (for some appropriate natural number \(\displaystyle 0 \le m < k/3\)).

Proposed by K.\(\displaystyle \,\)A. Kozma, Győr

(3 pont)

Deadline expired on March 10, 2023.

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

Megoldás. Mivel \(\displaystyle k \mid 1722-2m\) és \(\displaystyle k \mid 2179-3m\), ezért

\(\displaystyle k \mid 3 (1722-2m) - 2 (2179-3m) = 3 \cdot 1722 - 2 \cdot 2179 = 808. \)

A 808 prímtényezős felbontása \(\displaystyle 2^3 \cdot 101\), tehát az osztói csökkenő sorrendben: \(\displaystyle 808,404,202,101,8,4,2,1\).

808-cal osztva: \(\displaystyle 1722 = 2 \cdot 808 + 106\), míg \(\displaystyle 2179 = 2 \cdot 808 + 563\), itt \(\displaystyle \frac{106}{2} \neq \frac{563}{3}\), tehát nincs alkalmas \(\displaystyle m\).

404-gyel osztva \(\displaystyle 1722 = 4 \cdot 404 + 106\), míg \(\displaystyle 2179 = 5 \cdot 404 + 159\), itt \(\displaystyle m = 53\)-mal teljesül, hogy \(\displaystyle 2m = 106\) és \(\displaystyle 3m = 159\).

Tehát a válasz a feladat kérdésére: \(\displaystyle 404\) a legnagyobb ilyen egész szám.

Statistics:

118 students sent a solution.
3 points:	93 students.
2 points:	4 students.
1 point:	10 students.
0 point:	6 students.
Unfair, not evaluated:	2 solutionss.

Problems in Mathematics of KöMaL, February 2023