Problem K. 666. (October 2020)
K. 666. How many six-digit multiples of 182 are there in which the three-digit number formed by the first three digits is equal to the three-digit number formed by the last three digits?
(6 pont)
Deadline expired on November 10, 2020.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. A hatjegyű számot \(\displaystyle \overline{abcabc}\) alakban keressük. A feladat szövege szerint:
\(\displaystyle \overline{abcabc}=182\cdot n~\textrm{(ahol \(\displaystyle n\) egész szám)}.\)
\(\displaystyle \overline{abc}\cdot1001=182\cdot n,\)
\(\displaystyle \overline{abc}\cdot7\cdot11\cdot13=2\cdot7\cdot13\cdot n,\)
\(\displaystyle \overline{abc}\cdot11=2n.\)
Mivel \(\displaystyle 2n\) páros szám, így \(\displaystyle \overline{abc}\) is páros szám. \(\displaystyle \overline{abc}\) legkisebb értéke \(\displaystyle 100\) (ekkor \(\displaystyle n = 550\)). Legnagyobb értéke \(\displaystyle 998\) lehet (ekkor \(\displaystyle n = 5489\)). A legkisebb ilyen hatjegyű szám a \(\displaystyle 100\,100\), a legnagyobb ilyen hatjegyű szám a \(\displaystyle 998\,998\) és összesen \(\displaystyle \frac{998-100}{2}+1=450\) megfelelő szám van.
Statistics:
158 students sent a solution. 6 points: 76 students. 5 points: 21 students. 4 points: 9 students. 3 points: 8 students. 2 points: 9 students. 1 point: 17 students. 0 point: 12 students. Unfair, not evaluated: 6 solutionss.
Problems in Mathematics of KöMaL, October 2020