 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.

### Statistics:

 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