Notam cele trei numere astfel:
a+b+c=7925
Din fiecare numar scadem 975.
Se obtin resturi:
r.a=r.b+r.c (rest a=rest b+rest c)
r.b=a÷2
Si avem asa:
a-975=?
b-975=?
c-975=?
Dar stim ca:a+b+c=7925
975×3=2925
7925-2925=5000
r.c/__/
r.b/__/
r.a/__/__/
5000÷4=1250(b si c)
1250×2=2500(a)
Proba:1250+975+1250+975+2500+975=7925