Problem K. 361. (January 2013)
K. 361. A digital clock displays twelve noon in the form 12:00. Let us interpret that as a four-digit number: 1200. The intermediate times (displayed as hours and minutes) between twelve noon and midnight can all be interpreted in the same way as four-digit numbers. What is the sum of all such readings of four-digit numbers from 12:00 to 23:59?
(6 pont)
Deadline expired on February 11, 2013.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Írjuk fel óránként a megfelelő összeget:
\(\displaystyle 1200 + 1201 + 1202 + ... + 1258 + 1259 + \)
\(\displaystyle +1300 + 1301 + 1302 + ... + 1358 + 1359 + \)
\(\displaystyle +1400 + 1401 + 1402 + ... + 1458 + 1459 + \)
\(\displaystyle ...\)
\(\displaystyle +2300 + 2301 + 2302 + ... + 2358 + 2359.\)
Mind a 12 sorban szerepel az \(\displaystyle 1+2+3+...+58+59\) összeg. Ez összesen:
\(\displaystyle 12\cdot(1+2+3+...+58+59)= 12\cdot\frac{60\cdot59}{2} =21240.\)
Mind a 60 oszlopban szerepel az \(\displaystyle 1200+1300+...+2200+2300\) összeg. Ez összesen:
\(\displaystyle 60\cdot100\cdot(12+13+...+22+23) = 6000\cdot\frac{35\cdot12}{2} = 1260000. \)
Más tag nincs az összegben, tehát a keresett összeg: \(\displaystyle 21 240 + 1 260 000 = 1 281 240\).
Statistics:
165 students sent a solution. 6 points: 76 students. 5 points: 22 students. 4 points: 15 students. 3 points: 11 students. 2 points: 6 students. 1 point: 21 students. 0 point: 14 students.
Problems in Mathematics of KöMaL, January 2013