Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem C. 830. (December 2005)

C. 830. Lord Moneybag said to his grandson, ``Listen, Bill. Christmas is coming up. I have got some money between 300 and 500 pounds on me. It is an integer multiple of 6 pounds. I will give you 5 pounds out of it in one-pound coins. As I hand you the coins one by one, the amount that remains on me will first be divisible by 5, then by 4, then by 3, then by 2 and finally only by 1. If you can tell me how many pounds I have got on me, you'll get an extra ten-pound note.'' How much money did the lord have on him?

(5 pont)

Deadline expired on January 16, 2006.

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

Megoldás: Jelölje f az összeget. Tudjuk, hogy f=6k (ahol k egész). Ekkor biztos, hogy 3|f-3 és 2|f-4. Mivel 4|f-2=6k-2, azért f=12l+6 (ahol l egész). Mivel 5|f-1=12l+5, és (5,12)=1, azért 5|l. Vagyis f=12.5m+6=60m+6 (ahol m egész). Azt is tudjuk, hogy f-5=60m+1 prímszám, és 295 és 495 közé esik. Ilyen szám csak egy van, a 421. Tehát a Lordnál 426 font volt.


492 students sent a solution.
5 points:302 students.
4 points:115 students.
3 points:27 students.
2 points:16 students.
1 point:13 students.
0 point:13 students.
Unfair, not evaluated:6 solutionss.

Problems in Mathematics of KöMaL, December 2005