Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# 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